Euclid. # Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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Online Library → Euclid → Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools → online text (page 34 of 38)

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given angle.

95. Let A, B be the two given points, join AB and bisect AB in C,

and draw CD perpendicular to AB, then the center of the required circle

will be in CD. From O the center of the iiiven circle draw CFG parallel

to CD, and meeting the circle in F and AB produced in G. At F draw

a chord FF' equal to the given chord. Then the circle which passes

through the points at B and F, passes also through F'.

96. Let the straight line joining the centers of the two circles be

produced both ways to meet the circumference of the exterior circle.

97. Let A be the common center of two circks, and BCDE the chord

such that BE is double of CD. From A, B draw AF, BG perpendicular

to BE. Join AC, and produce it to meet BG in G. Then AC may be

shewn to be equal to CG, and the angle CBG being a right angle, is the

angle in the semicircle described on CG as its diameter.

98. The lines joining the common center and the extremities of the

chords of the circles, may be shewn to contain unequal angles, and the

angles at the centers of the circles are double the angles at the circum-

ferences, it follows that the segments containing these unequal angles

are not similar.

99. Let AB, AC be the straight lines drawn from A, a point m

ON BOOK 111. SSS

the outer circle to touch the inner circle in the points D, E, and meet

the outer circle again at B, C. Join BC, DE. Prove BC double of DE.

Let O be the center, and draw the common diameter AOG inter-

secting BC in E, and join EE. Then the figure DBFE may be proved

to be a parallelogram.

100. This appears from Euc. iii. 14.

101. The given point may be either within or without the circle.

Draw a chord in the circle equal to the given chord, and describe a

concentric circle touching the chord, and through the given point draw

a line touching this latter circle.

102. The diameter of the inner circle must not be less than one- third

of the diameter of the exterior circle.

103. Suppose AD, DB to be the tangents to the circle AEB contain-

ing the given angle. Draw DC to the center C and join CA, CB.

Then the triangles ACD, BCD are always equal : DC bisects the given

angle at D and the angle ACB. The angles CAB, CBD, being right

angles, are constant, and the angles ADC, BDC are constant, as also the

angles ACD, BC/D ; also AC, CB the radii of the given circle. Hence

the locus of D is a circle whose center is C and radius CD.

104. Let C be the center of the inner circle ; draw any radius CD,

at D draw a tangent CE equal to CD, joi\i CE, and with center C and

radius CE describe a circle and produce ED to meet the circle again in F.

lOo. Take C the center of the given circle, and draw any radius CD,

at D draw DE perpendicular to DC and equal to the length of the re-

quired tangent ; with center C and radius CE describe a circle,

106. This is manifest from Euc. iii. 36.

107. Let AB, AC be the sides of a triangle ABC. From A draw

the perpendicular AD on the opposite side, or opposite side produced,

The semicircles described on AB, BC both pass through D. Euc. iii. 31 .

108. Let A be the right angle of the triangle ABC, the first property

follows from the preceding Theorem 107. Let DE, DF be drawn to E,

F the centers of the circles on AB, AC and join EF. Then ED may

be proved to be perpendicular to the radius DF of the circle on AC at

the point D.

109. Let ABC be a triangle, and let the arcs be described on the

sides externally containing angles, whose sum is equal to two right angle.^.

It is obvious that the sum of the angles in the remaining segments is

equal to four right angles. These arcs may be shewn to intersect each

other in one point D. Let a, b, c be the centers of the circles on BC,

AC, AB. Join ab, be, ca ; Ab, bC, CÂ« ; aB, Be, cA ; bJ), cD, oD. Then

the angle cba may be proved equal to one-half of the angle AbC,

Similarly, the other two angles of abc.

110. It may be remarked, that generally, the mode of proof by which,

in pure geometry, three lines must, under specified conditions, pass

through the same point, is that by reductio ad absurdum. This will for

the most part require the converse theorem to be first proved or taken

for granted.

The converse theorem in this instance is, "If two perpendiculars

drawn from two angles of a triangle upon the opjjosite sides, intersect

in a ptnnt, the line drawn from the third angle through this point

will be perpendicular to the third side."

The proof will be formally thus : Let EHD be the triangle, AC,

BD two perpendiculars intersecting in F. If the third perpendicular

EG do not pass through F, let it take some other position as EH ; and

through F draw EFG to meet AD in G. Then it has been proved that

334 GEOMETRICAL EXERCISES, &C.

EG is perpendicular to AD : whence the two angles EHG, EGH of the

triangle EGH are equal to two right angles : â€” which is absurd.

111. The circle described on AB as a diameter will pass through

E and D. Then Euc. in. 36.

112. Since all the triangles are on the same base and have equal

vertical angles, these angles are in the same segment of a given circle.

The lines bisecting the vertical angles may be shewn to pass through

the extremity of that diameter which bisects the base.

113. Let AC be the common base of the triangles, ABC the isosceles

triangle, and ADC any other triangle on the same base AC and be-

tween the same parallels AC, BD. Describe a circle about ABC, and

let it cut AD in E and join EC. Then, Euc. i. 17, m. 21.

114. Let ABC be the given isosceles triangle having the vertical

angle at C, and let FG be anv given line. Required to find a point P

in FG such that the distance PA shall be double of PC. Divide AC

in D so that AD is double of DC, produce AC to E and make AE double

of AC. On DE describe a circle cutting FG in P, then PA is double

of PC. This is found by shewing that AP^ = 4 . PC^

115. On any two sides of the triangle, describe segments of circles

each containing an angle equal to two-thirds of a right angle, the point

of intersection of the arcs within the triangle will be the point required,

such that three lines drawn from it to the angles of the triangle shall

contain equal angles. Euc. in. 22.

116. Let A be the base of the tower, AB its altitude BC the height

of the flagstaff, AD a horizontal line drawn from A. If a circle be des-

cribed passing through the points B, C, and touching the line AD in

the point E : E will be the point required. Give the analysis.

117. If the ladder be supposed to be raised in a vertical plane, the

locus of the middle point may be shewn to be a quadrantal arc of which

the radius is half the length of the ladder.

118. The line drawn perpendicular to the diameter from the other

extremity of the tangent is parallel to the tangent drawn at the extremity

of the diameter.

119. Apply Euc. III. 21.

120. Let A, B, C, be the centers of the three equal circles, and let

them intersect one another in the point D : and let the circles whose

centers are A, B intersect each other again in E ; the circles whose cen-

ters are B, C in F ; and the circles whose centers are C, A in G. Then

FG is perpendicular to DE ; DG to FC ; and DF to GE. Since the

circles are equal, and all pass through the same point D, the centers A,

B, C are in a circle about D whose radius is the same as the radius of

the given circles. Join AB, BC, CA ; then these will be perpendicular

to the chords DE, DF, DG. Again, the figures DAGC, DBFC, are

equilateral, and hence FG is parallel to AB ; that is, perpendicular to

DE. Similarly for the other two cases.

121. Let E be the center of the circle which touches the two equal

circles whose centers are A, B. Join AE, BE which pass through the

points of contact F, G. Whence AE is equal to EB. Also CD the

common chord bisects AB at right angles, and therefore the perpen-

dicular from E on AB coincides with CD.

122. Let three circles touch each other at the point A, and from A

let a line ABCD be drawn cutting the circumferences in B, C, D. Let

O, O', O" be the centers of the circles, join BO, CO', DO", these lines

are parallel to one another. Euc. i. 6. 28.

123. Proceed as in Theorem 110, supra.

I

ON BOOK III. 335

1?4. The three tangents will be found to be perpendicular to the

sides of the triangle formed by joining the centers of the three circles.

125. With center A and any radius less than the radius of either of

the equal circles, describe the third circle intersecting them in C and D.

Join EC, CD, and prove EC and CD to be in the same straight line.

126. Let AEC be the triangle required ; EC the given base, ED the

given difference of the sides, and BAC the given vertical angle. Join

CD and draw AM perpendicular to CD. Then MAD is half the vertical

angle and AMD a right angle : the angle EDC is therefore given, and

hence D is a point in the arc of a given segment on EC. Also since ED

is given, the point D is given, and therefore the sides BA, AC are given.

Hence the synthesis.

127. Let AEC be the required triangle, AD the line bisecting the

vertical angle and dividing the base EC into the segments ED, DC.

About the triangle AEC describe a circle and produce AD to meet the

circumference in E, then the arcs BE, EC are equal.

128. Analysis. Let ABC be the triangle, and let the circle ABC be

described about it : draw AF to bisect the vertical angle BAC and meet

the circle in F, make AV equal to AC, and draw CV to meet the circle

in T ; join TB and TF, cutting AB in D ; draw the diameter FS cutting

BC in R, DR cutting AF in E ; join AS, and draw AK, AH perpen-

dicular to FS and EC. Then shew that AD is half the sum, and DB

half the difference of the sides AB, AC. Next, that the point F in which

AF meets the circumscribing circle is given, also the point E where DE

meets AF is given. The points A, K, K, E are in a circle, Euc. iii. 22.

Hence, KF . FR = AF . FE, a given rectangle ; and the segment KR,

which is equal to the perpendicular AH, being given, RF itself is given.

"Whence the construction,

129. On AB the given base describe a circle such that the segment

AEB shall contain an angle equal to the given vertical angle of the tri-

angle. Draw the diameter EMD cutting AB in M at right angles. At

D in ED, make the angle EDC equal to half the given difference of the

angles at the base, and let DC meet the circumference of the circle in C.

Join CA, CB ; ABC is the triangle required. For, make CF equal to

CB, and join FE cutting CD in G.

130. 'Let AEC be the triangle, AD the perpendicular ori BC. With

center A, and AC the less side as radius, describe a circle cutting the

base BC in E, and the longer side AB in G, and BA produced in F, and

join AE, EG, FC. Then the angle GFC being half the given angle,

EAC is given, and the angle BEG equal to GFC is also given. Like-

wise BE the difference of the segments of the base, and EG the difference

of the sides, are given by the problem. Wherefore the triangle BEG is

given (with two solutions). Again, the angle EGB being given, the

angle AGE, and hence its equal AEG is given ; and hence the vertex A

is given, and likewise the line AE equal to AC the shortest side is given.

Hence the construction.

131. Let AEC be the triangle, D, E the bisections of the sides AC,

AB. Join CE, ED intersecting in F. Bisect ED in G and join EG. Then

EF, one-third of EC is given, and EG one-half of ED is also given.

Now EG is parallel to AC ; and the angle BAC being given, its equal

opposite angle BEG is also given. Whence the segment of the circle

containing the angle BEG is also given. Hence F is a given point, and

FE a given line, whence E is in the circumference of the given circle

about F whose radius is FE. Wherefore E being in two given circles, it,

is itself their given intersection.

336 GEOMETRICAL EXERCISES, &C.

132. Of all triangles on the same base and having equal vertical

angles, that triangle will be the greatest whose perpendicular from the

vertex on the base is a maximum, and the greatest perpendicular is that

which bisects the base. Whence the triangle is isosceles.

133. Let AB be the given base and ABC the sum of the other two

sides ; at B draw BD at right angles to AB and equal to the given alti-

tude, produce BD to E making DE equal to BD. With center A and

with radius AC describe the circle CFG, draw FO at right angles to BE

and find in it the center O of the circle which passes through B and B

and touches the former circle in the point F. The centers A, being

joined and the line produced, will pass through F, Join OB. Then

AOB is the triangle required.

134. Since the area and bases of the triangle are given, the altitude

is given. Hence the problem is â€” given- the base, the vertical angle and

the altitude, describe the triangle.

135. Apply Euc. iii. 27.

136. The fixed point may be proved to be the center of the circle.

137. Let the line which bisects any angle BAD of the quadrilateral,

meet the circumference in E, join EC, and prove that the angle made by

producing DC is bisected by EC.

138. Draw the diagonals of the quadrilateral, and by Euc. iii. 21 , i. 29.

139. From the center draw lines to the angles : then Euc. iii. 27.

140. The centers of the four circles are determined by the intersec-

tion of the lines which bisect the four angles of the given quadrilateral.

Join these four points, and the opposite angles of the quadrilateral so

formed are respectively equal to two right angles.

141 . Let ABCD be the required trapezium inscribed in the given circle

(fig. Euc. III. 22.) of which AB is given, also the sum of the remaining

three sides and the angle ADC. Since the angle ADC is given, the

opposite angle ABC is known, and therefore the point C and the side

BC. Produce AD and make DE equal to DC and join EC. Since the

sum of AD, DC, CB is given, and DC is known, therefore the sum of

AD, DC is given, and likewise AC, and the angle ADC. Also the angle

DEC being half of the angle ADC is given. Whence the segment of the

circle which 'contains AEC is given, also AE is given, and hence the

point E, and consequently the point D. Whence the construction.

142. Let ADBC be the inscribed quadrilateral ; let AC, BD pro-

duced meet in O, and AB, CD produced meet in P, also let the tangents

from 0, P meet the circles in K, H respectively. Join OP, and about

the triangle PAC describe a circle cutting PO in G and join AG. Then

A, B, G, O may be shewn to be points in the circumference of a circle.

Whence the sum of the squares on OH and PK may be found by Euc.

III. 36, and shewn to be equal to the square on OP.

143. This will be manifest from the equaUty of the two tangents

drawn to a circle from the same poiut.

144. Apply Euc. iii. 22.

145. A circle can be described about the figure AECBF.

146. Apply Euc. iii. 22, 32.

147. Apply Euc. III. 21, 22, 32.

148. Apply Euc. iii. 20, and the angle BAD will be found to be

double of the angles CBD and CDB together.

149. Let ABCD be the given quadrilateral figure, and let the angles

at A, B, C, D be bisected by four lines, so that the lines which bisect the

angles A and B, B and C, C and D, D and A, meet in the points a, h, c, d,

respectively. Prove that the angles at a and c, or at h and (Z, are to-

gether equal to two right angles.

ON BOOK III. obi

loO. Apply Euc. III. 22.

151. Join the center of the circle with the other extremity of the line

perpendicular to the diameter.

152. Let AB be a chord parallel to the diameter FG of the circle,

fig. Theo. 1, p. 160, and H any point in the diameter. Let HA and HB

be join-ed. Bisect FG in O, draw OL perpendicular to FG cutting AB

in K, and join HK, IlL, OA. Then the square on HA and HF may be

proved equal to the squares on FH, HG by Theo. 3, p. 114 ; Euc. i. 47;

Euc, II. 9.

153. Let A be the given point (fig. Euc. in. 36, Cor.) and suppose

AFC meeting the circle in F, C, to be bisected in F, and let AD be a

tangent drawn from A. Then 2. AF^ = AF . AC = AD', but AD is

given, hence also AF is given. To construct. Draw the tangent AD.

On AD describe a semicircle AGD, bisect it in G ; with center A and

radius AG, describe a circle cutting the given circle in F. Join AF and

produce it to meet the circumference again in C.

154. Let the chords AB, CD intersect each other in E at right

angles. Find F the center, and draw the diameters HEFG, AFK and

join AC, CK, BD. Then by Euc. ii. 4. 5 ; in. 35.

155. Let E, F be the points in the diameter AB equidistant from the

center O ; CED any chord; draw OG perpendicular to CED, and join

FG, OC. The sum of the squares on DF and FC may be shewn to be

equal to twice the square on FE and the rectangle contained by AE, EB

by Euc. I. 47 ; II. 5 ; ni. 35.

156. Let the chords AB, AC be drawn from the point A, and let a

chord FG parallel to the tangent at A be drawn intersecting the chords

AB, AC in D and E, and join BC. Then the opposite angles of the

quadrilateral BDEC are equal to two right angles, and a circle would

circumscribe the figure. Hence by Euc. i. 36.

157. Let the lines be drawn as directed in the enunciation. Draw

the diameter AE and join CE, DE, BE ; then AC^ + AD^ and 2 . AB'

may be each shewn to be equal to the square on the diameter.

158. Let QOP cut the diameter AB in O. From C the center draw

CH perpendicular to QP. Then CH is equal to OH, and by Euc. n. 9,

the squares on PO, OQ are readily shewn to be equal to twice the square

onCP.

159. From P draw PQ perpendicular on AB meeting it in Q. Join

AC, CD, DB. Then circles would circumscribe the quadrilaterals ACPQ

and BDPQ, and then by Euc. in. 36.

160. Describe the figure according to the enunciation ; draw AE the

diameter of the circle, and let P be the intersection Â©â– f the diagonals of the

parallelogram. Draw EB, EP, EC, EF, EG, EH. Since AE is a

diameter of the circle, the angles fit F, G, H, are right angles, and EF,

EG, EH are perpendiculars from the vortex upon the bases of the tri-

angles EAB, EAC, EAP. Whence by Euc. ii. 1 3, and theorem 3, page

114, the truth of the property may be shewn.

161. If FA be the given line (fig. Euc. ii. 11), and if FA be produced

to C ; AC is the part produced which satisfies the required conditions.

162. Let AD meet the circle in G, H, and join BG, GC. ThenBGC

is a right-angled triangle and GD is perpendicular to the hypotenuse,

and the rectangles may be each shewn to be equal to the square on BG.

Euc. in. 35 ; ii. 5 ; i. 47. Or, if EC be joined, the quadrilateral

figure ADCE may be circumscribed by a circle. Euc. in. 31, 22, 36, Cor.

163. On PC describe a semicircle cutting the given one in E, and

draw EF perpendicular to AD ; then F is the point required,

Q

338 GEOMETRICAL EXERCISES, &C.

164. Let AB be the given straight line. Bisect AB in C and on AB

as a diameter describe a circle ; and at any point D in the circumference,

draw a tangent DE equal to a side of the given square ; join DC, EC,

and with center C and radius CE describe a circle cutting AB produced

in F. From F draw FG to touch the circle whose center is C in

the point G.

165. Let AD, DF be two lines at right angles to each other, O the

centre of the circle BFQ ; A any point in AD from which tangents AB,

AC are drawn; then the chord BC shall always cut FD in the same

point P, wherever the point A is taken in AD. Join AP ; then BAC is

an isosceles triangle,

and FD . DE + AD^ = AB^ = BP . PC + AP^ = BP . PC + AD^ + DP^

wherefore BP . PC = FD . DE - DP^

The point P, therefore, is independent of the position of the point A ; and

is consequently the same for all positions of A in the line AD.

166. The point E will be found to be that point in BC, from which

two tangents to the circles described on AB and CD as diameters, are

equal, Euc. in. 36.

167. If AQ, A'P' be produced to meet, these lines with AA' form a

right-angled triangle, then Euc. i. 47.

GEOMETRICAL EXERCISES ON BOOK IV.

HINTS, &c.

1. Let AB be the given line. Draw through C the center of the

given circle the diameter DCE. Bisect AB in F and join FC, Through

A, B draw AG, BH parallel to FC and meeting the diameter in G, H :

at G, H draw GK, HL perpendicular to DE and meeting the circumfer-

ence in the points K, L ; join KL; then KL is equal and parallel to AB.

2. Trisect the circumference and join the center with the points of

trisection.

3. See Euc. iv. 4, 5.

4. Let a line be drawn from the third angle to the point of intersec-

tion of the two lines ; and the three distances of this point from the angles

may be shewn to be equal.

5. Let the line AD drawn from the vertex A of the equilateral tri-

angle, cut the base BC, and meet the circumference of the circle in D.

Let DB, DC be joined : AD is equal to DB and DC. If on DA, DE be

taken equal to DB, and BE be joined ; BDE may be proved to be an

equilateral triangle, also the triangle ABE may be proved equal to the

triangle CBD.

The other case is when the lirTe does not cut the base.

6. Let a circle be described upon the base of the equilateral triangle,

and let an equilateral triangle be inscribed in the circle. Draw a diameter

from one of the vertices of the inscribed triangle, and join the other ex-

tremity of the diameter with one of the other extremities of the sides of

the inscribed triangle. The side of the inscribed triangle may then te

proved to be equal to the perpendicular in the other triangle.

7. The line joining the points of bisection, is parallel to the base of

the triangle and therefore cuts off an equilateral triangle from the given

triangle. By Euc. iii. 21 ; i. 6, the truth of the theorem may be shewn.

8. Let a diameter be drawn from any angle of an equilateral tri-

ON BOOK IV. 339

angle inscribed in a circle to meet the circumference. It may be proved

that the radius is bisected by the opposite side of the triangle.

9. Let ABC be an equilateral triangle inscribed in a circle, and let

AB'C be an isosceles triangle inscribed in the same circle, having the

same vertex A. Draw the diameter AD intersecting BC in E, and B'C

in E', and let B'C fall below BC. Then AB, BE, and AB', B'E', are

respectively the semi-perimeters of the triangles. Draw B'F perpendi-

cular to BC, and cut off AH equal to AB, and join BH. If BF can be

proved to be greater than B'H, the perimeter of ABC is greater than the

perimeter of AB'C. Next let B C fall above BC.

10. The angles contained in the two segments of the circle, may be

shewn to be equal, then by joining the extremities of the arcs, the two

remaining sides may be shewn to be parallel.

11. It may be shewn that four equal and equilateral triangles will

form an equilateral triangle of the same perimeter as the hexagon, which

is formed by six equal and equilateral triangles.

12. Let the figure be constructed. By drawing the diagonals of the

hexagon, the proof is obvious.

13. By Euc. I. 47, the perpendicular distance from the center of the

circle upon the side of the inscribed hexagon may be found.

14. The alternate sides of the hexagon will fall upon the sides of the

:riangle, and each side will be found to be equal to one-third of the side

Df the equilateral triangle.

15. A regular duodecagon may be inscribed in a circle by means of

the equilateral triangle and square, or by means of the hexagon. The

irea of the duodecagon is three times the square on the radius of the circle,

which is the square on the side of an equilateral triangle inscribed in the

same circle. Theorem 1, p. 196.

16. In general, three straight lines when produced will meet and

hrm. a triangle, except when all three are parallel or two parallel are

ntersected by the third. This Problem includes Euc. iv. 5, and all the

ases which arise from producing the sides of the triangle. The circles

lescribed touching a side of a triangle and the other two sides produced,

ire called the escribed circles.

17. This is manifest from Euc. iii. 21.

18. The point required is the center of the circle which circumscribes

;he triangle. See the notes on Euc. iii. 20, p. 155.

19. if the perpendiculars meet the three sides of the triangle, the

)oint is within the triangle, Euc. iv. 4. If the perpendiculars meet the

)ase and the two sides produced, the point is the center of the escribed

ircle.

20. This is manifest from Euc. in. 11, 18.

21. The base BC is intersected by the perpendicular AD, and the

ide AC is intersected by the perpendicular BE. From Theorem i. p.

60 ; the arc AF is proved equal to AE, or the arc FE is bisected in A.

!n the same manner the arcs FD, DE, may be shewn to be bisected inBC.

22. Let ABC be a triangle, and let D, E be the points where the in-

cribed circle touches the sides AB, AC. Draw BE, CD intersecting

95. Let A, B be the two given points, join AB and bisect AB in C,

and draw CD perpendicular to AB, then the center of the required circle

will be in CD. From O the center of the iiiven circle draw CFG parallel

to CD, and meeting the circle in F and AB produced in G. At F draw

a chord FF' equal to the given chord. Then the circle which passes

through the points at B and F, passes also through F'.

96. Let the straight line joining the centers of the two circles be

produced both ways to meet the circumference of the exterior circle.

97. Let A be the common center of two circks, and BCDE the chord

such that BE is double of CD. From A, B draw AF, BG perpendicular

to BE. Join AC, and produce it to meet BG in G. Then AC may be

shewn to be equal to CG, and the angle CBG being a right angle, is the

angle in the semicircle described on CG as its diameter.

98. The lines joining the common center and the extremities of the

chords of the circles, may be shewn to contain unequal angles, and the

angles at the centers of the circles are double the angles at the circum-

ferences, it follows that the segments containing these unequal angles

are not similar.

99. Let AB, AC be the straight lines drawn from A, a point m

ON BOOK 111. SSS

the outer circle to touch the inner circle in the points D, E, and meet

the outer circle again at B, C. Join BC, DE. Prove BC double of DE.

Let O be the center, and draw the common diameter AOG inter-

secting BC in E, and join EE. Then the figure DBFE may be proved

to be a parallelogram.

100. This appears from Euc. iii. 14.

101. The given point may be either within or without the circle.

Draw a chord in the circle equal to the given chord, and describe a

concentric circle touching the chord, and through the given point draw

a line touching this latter circle.

102. The diameter of the inner circle must not be less than one- third

of the diameter of the exterior circle.

103. Suppose AD, DB to be the tangents to the circle AEB contain-

ing the given angle. Draw DC to the center C and join CA, CB.

Then the triangles ACD, BCD are always equal : DC bisects the given

angle at D and the angle ACB. The angles CAB, CBD, being right

angles, are constant, and the angles ADC, BDC are constant, as also the

angles ACD, BC/D ; also AC, CB the radii of the given circle. Hence

the locus of D is a circle whose center is C and radius CD.

104. Let C be the center of the inner circle ; draw any radius CD,

at D draw a tangent CE equal to CD, joi\i CE, and with center C and

radius CE describe a circle and produce ED to meet the circle again in F.

lOo. Take C the center of the given circle, and draw any radius CD,

at D draw DE perpendicular to DC and equal to the length of the re-

quired tangent ; with center C and radius CE describe a circle,

106. This is manifest from Euc. iii. 36.

107. Let AB, AC be the sides of a triangle ABC. From A draw

the perpendicular AD on the opposite side, or opposite side produced,

The semicircles described on AB, BC both pass through D. Euc. iii. 31 .

108. Let A be the right angle of the triangle ABC, the first property

follows from the preceding Theorem 107. Let DE, DF be drawn to E,

F the centers of the circles on AB, AC and join EF. Then ED may

be proved to be perpendicular to the radius DF of the circle on AC at

the point D.

109. Let ABC be a triangle, and let the arcs be described on the

sides externally containing angles, whose sum is equal to two right angle.^.

It is obvious that the sum of the angles in the remaining segments is

equal to four right angles. These arcs may be shewn to intersect each

other in one point D. Let a, b, c be the centers of the circles on BC,

AC, AB. Join ab, be, ca ; Ab, bC, CÂ« ; aB, Be, cA ; bJ), cD, oD. Then

the angle cba may be proved equal to one-half of the angle AbC,

Similarly, the other two angles of abc.

110. It may be remarked, that generally, the mode of proof by which,

in pure geometry, three lines must, under specified conditions, pass

through the same point, is that by reductio ad absurdum. This will for

the most part require the converse theorem to be first proved or taken

for granted.

The converse theorem in this instance is, "If two perpendiculars

drawn from two angles of a triangle upon the opjjosite sides, intersect

in a ptnnt, the line drawn from the third angle through this point

will be perpendicular to the third side."

The proof will be formally thus : Let EHD be the triangle, AC,

BD two perpendiculars intersecting in F. If the third perpendicular

EG do not pass through F, let it take some other position as EH ; and

through F draw EFG to meet AD in G. Then it has been proved that

334 GEOMETRICAL EXERCISES, &C.

EG is perpendicular to AD : whence the two angles EHG, EGH of the

triangle EGH are equal to two right angles : â€” which is absurd.

111. The circle described on AB as a diameter will pass through

E and D. Then Euc. in. 36.

112. Since all the triangles are on the same base and have equal

vertical angles, these angles are in the same segment of a given circle.

The lines bisecting the vertical angles may be shewn to pass through

the extremity of that diameter which bisects the base.

113. Let AC be the common base of the triangles, ABC the isosceles

triangle, and ADC any other triangle on the same base AC and be-

tween the same parallels AC, BD. Describe a circle about ABC, and

let it cut AD in E and join EC. Then, Euc. i. 17, m. 21.

114. Let ABC be the given isosceles triangle having the vertical

angle at C, and let FG be anv given line. Required to find a point P

in FG such that the distance PA shall be double of PC. Divide AC

in D so that AD is double of DC, produce AC to E and make AE double

of AC. On DE describe a circle cutting FG in P, then PA is double

of PC. This is found by shewing that AP^ = 4 . PC^

115. On any two sides of the triangle, describe segments of circles

each containing an angle equal to two-thirds of a right angle, the point

of intersection of the arcs within the triangle will be the point required,

such that three lines drawn from it to the angles of the triangle shall

contain equal angles. Euc. in. 22.

116. Let A be the base of the tower, AB its altitude BC the height

of the flagstaff, AD a horizontal line drawn from A. If a circle be des-

cribed passing through the points B, C, and touching the line AD in

the point E : E will be the point required. Give the analysis.

117. If the ladder be supposed to be raised in a vertical plane, the

locus of the middle point may be shewn to be a quadrantal arc of which

the radius is half the length of the ladder.

118. The line drawn perpendicular to the diameter from the other

extremity of the tangent is parallel to the tangent drawn at the extremity

of the diameter.

119. Apply Euc. III. 21.

120. Let A, B, C, be the centers of the three equal circles, and let

them intersect one another in the point D : and let the circles whose

centers are A, B intersect each other again in E ; the circles whose cen-

ters are B, C in F ; and the circles whose centers are C, A in G. Then

FG is perpendicular to DE ; DG to FC ; and DF to GE. Since the

circles are equal, and all pass through the same point D, the centers A,

B, C are in a circle about D whose radius is the same as the radius of

the given circles. Join AB, BC, CA ; then these will be perpendicular

to the chords DE, DF, DG. Again, the figures DAGC, DBFC, are

equilateral, and hence FG is parallel to AB ; that is, perpendicular to

DE. Similarly for the other two cases.

121. Let E be the center of the circle which touches the two equal

circles whose centers are A, B. Join AE, BE which pass through the

points of contact F, G. Whence AE is equal to EB. Also CD the

common chord bisects AB at right angles, and therefore the perpen-

dicular from E on AB coincides with CD.

122. Let three circles touch each other at the point A, and from A

let a line ABCD be drawn cutting the circumferences in B, C, D. Let

O, O', O" be the centers of the circles, join BO, CO', DO", these lines

are parallel to one another. Euc. i. 6. 28.

123. Proceed as in Theorem 110, supra.

I

ON BOOK III. 335

1?4. The three tangents will be found to be perpendicular to the

sides of the triangle formed by joining the centers of the three circles.

125. With center A and any radius less than the radius of either of

the equal circles, describe the third circle intersecting them in C and D.

Join EC, CD, and prove EC and CD to be in the same straight line.

126. Let AEC be the triangle required ; EC the given base, ED the

given difference of the sides, and BAC the given vertical angle. Join

CD and draw AM perpendicular to CD. Then MAD is half the vertical

angle and AMD a right angle : the angle EDC is therefore given, and

hence D is a point in the arc of a given segment on EC. Also since ED

is given, the point D is given, and therefore the sides BA, AC are given.

Hence the synthesis.

127. Let AEC be the required triangle, AD the line bisecting the

vertical angle and dividing the base EC into the segments ED, DC.

About the triangle AEC describe a circle and produce AD to meet the

circumference in E, then the arcs BE, EC are equal.

128. Analysis. Let ABC be the triangle, and let the circle ABC be

described about it : draw AF to bisect the vertical angle BAC and meet

the circle in F, make AV equal to AC, and draw CV to meet the circle

in T ; join TB and TF, cutting AB in D ; draw the diameter FS cutting

BC in R, DR cutting AF in E ; join AS, and draw AK, AH perpen-

dicular to FS and EC. Then shew that AD is half the sum, and DB

half the difference of the sides AB, AC. Next, that the point F in which

AF meets the circumscribing circle is given, also the point E where DE

meets AF is given. The points A, K, K, E are in a circle, Euc. iii. 22.

Hence, KF . FR = AF . FE, a given rectangle ; and the segment KR,

which is equal to the perpendicular AH, being given, RF itself is given.

"Whence the construction,

129. On AB the given base describe a circle such that the segment

AEB shall contain an angle equal to the given vertical angle of the tri-

angle. Draw the diameter EMD cutting AB in M at right angles. At

D in ED, make the angle EDC equal to half the given difference of the

angles at the base, and let DC meet the circumference of the circle in C.

Join CA, CB ; ABC is the triangle required. For, make CF equal to

CB, and join FE cutting CD in G.

130. 'Let AEC be the triangle, AD the perpendicular ori BC. With

center A, and AC the less side as radius, describe a circle cutting the

base BC in E, and the longer side AB in G, and BA produced in F, and

join AE, EG, FC. Then the angle GFC being half the given angle,

EAC is given, and the angle BEG equal to GFC is also given. Like-

wise BE the difference of the segments of the base, and EG the difference

of the sides, are given by the problem. Wherefore the triangle BEG is

given (with two solutions). Again, the angle EGB being given, the

angle AGE, and hence its equal AEG is given ; and hence the vertex A

is given, and likewise the line AE equal to AC the shortest side is given.

Hence the construction.

131. Let AEC be the triangle, D, E the bisections of the sides AC,

AB. Join CE, ED intersecting in F. Bisect ED in G and join EG. Then

EF, one-third of EC is given, and EG one-half of ED is also given.

Now EG is parallel to AC ; and the angle BAC being given, its equal

opposite angle BEG is also given. Whence the segment of the circle

containing the angle BEG is also given. Hence F is a given point, and

FE a given line, whence E is in the circumference of the given circle

about F whose radius is FE. Wherefore E being in two given circles, it,

is itself their given intersection.

336 GEOMETRICAL EXERCISES, &C.

132. Of all triangles on the same base and having equal vertical

angles, that triangle will be the greatest whose perpendicular from the

vertex on the base is a maximum, and the greatest perpendicular is that

which bisects the base. Whence the triangle is isosceles.

133. Let AB be the given base and ABC the sum of the other two

sides ; at B draw BD at right angles to AB and equal to the given alti-

tude, produce BD to E making DE equal to BD. With center A and

with radius AC describe the circle CFG, draw FO at right angles to BE

and find in it the center O of the circle which passes through B and B

and touches the former circle in the point F. The centers A, being

joined and the line produced, will pass through F, Join OB. Then

AOB is the triangle required.

134. Since the area and bases of the triangle are given, the altitude

is given. Hence the problem is â€” given- the base, the vertical angle and

the altitude, describe the triangle.

135. Apply Euc. iii. 27.

136. The fixed point may be proved to be the center of the circle.

137. Let the line which bisects any angle BAD of the quadrilateral,

meet the circumference in E, join EC, and prove that the angle made by

producing DC is bisected by EC.

138. Draw the diagonals of the quadrilateral, and by Euc. iii. 21 , i. 29.

139. From the center draw lines to the angles : then Euc. iii. 27.

140. The centers of the four circles are determined by the intersec-

tion of the lines which bisect the four angles of the given quadrilateral.

Join these four points, and the opposite angles of the quadrilateral so

formed are respectively equal to two right angles.

141 . Let ABCD be the required trapezium inscribed in the given circle

(fig. Euc. III. 22.) of which AB is given, also the sum of the remaining

three sides and the angle ADC. Since the angle ADC is given, the

opposite angle ABC is known, and therefore the point C and the side

BC. Produce AD and make DE equal to DC and join EC. Since the

sum of AD, DC, CB is given, and DC is known, therefore the sum of

AD, DC is given, and likewise AC, and the angle ADC. Also the angle

DEC being half of the angle ADC is given. Whence the segment of the

circle which 'contains AEC is given, also AE is given, and hence the

point E, and consequently the point D. Whence the construction.

142. Let ADBC be the inscribed quadrilateral ; let AC, BD pro-

duced meet in O, and AB, CD produced meet in P, also let the tangents

from 0, P meet the circles in K, H respectively. Join OP, and about

the triangle PAC describe a circle cutting PO in G and join AG. Then

A, B, G, O may be shewn to be points in the circumference of a circle.

Whence the sum of the squares on OH and PK may be found by Euc.

III. 36, and shewn to be equal to the square on OP.

143. This will be manifest from the equaUty of the two tangents

drawn to a circle from the same poiut.

144. Apply Euc. iii. 22.

145. A circle can be described about the figure AECBF.

146. Apply Euc. iii. 22, 32.

147. Apply Euc. III. 21, 22, 32.

148. Apply Euc. iii. 20, and the angle BAD will be found to be

double of the angles CBD and CDB together.

149. Let ABCD be the given quadrilateral figure, and let the angles

at A, B, C, D be bisected by four lines, so that the lines which bisect the

angles A and B, B and C, C and D, D and A, meet in the points a, h, c, d,

respectively. Prove that the angles at a and c, or at h and (Z, are to-

gether equal to two right angles.

ON BOOK III. obi

loO. Apply Euc. III. 22.

151. Join the center of the circle with the other extremity of the line

perpendicular to the diameter.

152. Let AB be a chord parallel to the diameter FG of the circle,

fig. Theo. 1, p. 160, and H any point in the diameter. Let HA and HB

be join-ed. Bisect FG in O, draw OL perpendicular to FG cutting AB

in K, and join HK, IlL, OA. Then the square on HA and HF may be

proved equal to the squares on FH, HG by Theo. 3, p. 114 ; Euc. i. 47;

Euc, II. 9.

153. Let A be the given point (fig. Euc. in. 36, Cor.) and suppose

AFC meeting the circle in F, C, to be bisected in F, and let AD be a

tangent drawn from A. Then 2. AF^ = AF . AC = AD', but AD is

given, hence also AF is given. To construct. Draw the tangent AD.

On AD describe a semicircle AGD, bisect it in G ; with center A and

radius AG, describe a circle cutting the given circle in F. Join AF and

produce it to meet the circumference again in C.

154. Let the chords AB, CD intersect each other in E at right

angles. Find F the center, and draw the diameters HEFG, AFK and

join AC, CK, BD. Then by Euc. ii. 4. 5 ; in. 35.

155. Let E, F be the points in the diameter AB equidistant from the

center O ; CED any chord; draw OG perpendicular to CED, and join

FG, OC. The sum of the squares on DF and FC may be shewn to be

equal to twice the square on FE and the rectangle contained by AE, EB

by Euc. I. 47 ; II. 5 ; ni. 35.

156. Let the chords AB, AC be drawn from the point A, and let a

chord FG parallel to the tangent at A be drawn intersecting the chords

AB, AC in D and E, and join BC. Then the opposite angles of the

quadrilateral BDEC are equal to two right angles, and a circle would

circumscribe the figure. Hence by Euc. i. 36.

157. Let the lines be drawn as directed in the enunciation. Draw

the diameter AE and join CE, DE, BE ; then AC^ + AD^ and 2 . AB'

may be each shewn to be equal to the square on the diameter.

158. Let QOP cut the diameter AB in O. From C the center draw

CH perpendicular to QP. Then CH is equal to OH, and by Euc. n. 9,

the squares on PO, OQ are readily shewn to be equal to twice the square

onCP.

159. From P draw PQ perpendicular on AB meeting it in Q. Join

AC, CD, DB. Then circles would circumscribe the quadrilaterals ACPQ

and BDPQ, and then by Euc. in. 36.

160. Describe the figure according to the enunciation ; draw AE the

diameter of the circle, and let P be the intersection Â©â– f the diagonals of the

parallelogram. Draw EB, EP, EC, EF, EG, EH. Since AE is a

diameter of the circle, the angles fit F, G, H, are right angles, and EF,

EG, EH are perpendiculars from the vortex upon the bases of the tri-

angles EAB, EAC, EAP. Whence by Euc. ii. 1 3, and theorem 3, page

114, the truth of the property may be shewn.

161. If FA be the given line (fig. Euc. ii. 11), and if FA be produced

to C ; AC is the part produced which satisfies the required conditions.

162. Let AD meet the circle in G, H, and join BG, GC. ThenBGC

is a right-angled triangle and GD is perpendicular to the hypotenuse,

and the rectangles may be each shewn to be equal to the square on BG.

Euc. in. 35 ; ii. 5 ; i. 47. Or, if EC be joined, the quadrilateral

figure ADCE may be circumscribed by a circle. Euc. in. 31, 22, 36, Cor.

163. On PC describe a semicircle cutting the given one in E, and

draw EF perpendicular to AD ; then F is the point required,

Q

338 GEOMETRICAL EXERCISES, &C.

164. Let AB be the given straight line. Bisect AB in C and on AB

as a diameter describe a circle ; and at any point D in the circumference,

draw a tangent DE equal to a side of the given square ; join DC, EC,

and with center C and radius CE describe a circle cutting AB produced

in F. From F draw FG to touch the circle whose center is C in

the point G.

165. Let AD, DF be two lines at right angles to each other, O the

centre of the circle BFQ ; A any point in AD from which tangents AB,

AC are drawn; then the chord BC shall always cut FD in the same

point P, wherever the point A is taken in AD. Join AP ; then BAC is

an isosceles triangle,

and FD . DE + AD^ = AB^ = BP . PC + AP^ = BP . PC + AD^ + DP^

wherefore BP . PC = FD . DE - DP^

The point P, therefore, is independent of the position of the point A ; and

is consequently the same for all positions of A in the line AD.

166. The point E will be found to be that point in BC, from which

two tangents to the circles described on AB and CD as diameters, are

equal, Euc. in. 36.

167. If AQ, A'P' be produced to meet, these lines with AA' form a

right-angled triangle, then Euc. i. 47.

GEOMETRICAL EXERCISES ON BOOK IV.

HINTS, &c.

1. Let AB be the given line. Draw through C the center of the

given circle the diameter DCE. Bisect AB in F and join FC, Through

A, B draw AG, BH parallel to FC and meeting the diameter in G, H :

at G, H draw GK, HL perpendicular to DE and meeting the circumfer-

ence in the points K, L ; join KL; then KL is equal and parallel to AB.

2. Trisect the circumference and join the center with the points of

trisection.

3. See Euc. iv. 4, 5.

4. Let a line be drawn from the third angle to the point of intersec-

tion of the two lines ; and the three distances of this point from the angles

may be shewn to be equal.

5. Let the line AD drawn from the vertex A of the equilateral tri-

angle, cut the base BC, and meet the circumference of the circle in D.

Let DB, DC be joined : AD is equal to DB and DC. If on DA, DE be

taken equal to DB, and BE be joined ; BDE may be proved to be an

equilateral triangle, also the triangle ABE may be proved equal to the

triangle CBD.

The other case is when the lirTe does not cut the base.

6. Let a circle be described upon the base of the equilateral triangle,

and let an equilateral triangle be inscribed in the circle. Draw a diameter

from one of the vertices of the inscribed triangle, and join the other ex-

tremity of the diameter with one of the other extremities of the sides of

the inscribed triangle. The side of the inscribed triangle may then te

proved to be equal to the perpendicular in the other triangle.

7. The line joining the points of bisection, is parallel to the base of

the triangle and therefore cuts off an equilateral triangle from the given

triangle. By Euc. iii. 21 ; i. 6, the truth of the theorem may be shewn.

8. Let a diameter be drawn from any angle of an equilateral tri-

ON BOOK IV. 339

angle inscribed in a circle to meet the circumference. It may be proved

that the radius is bisected by the opposite side of the triangle.

9. Let ABC be an equilateral triangle inscribed in a circle, and let

AB'C be an isosceles triangle inscribed in the same circle, having the

same vertex A. Draw the diameter AD intersecting BC in E, and B'C

in E', and let B'C fall below BC. Then AB, BE, and AB', B'E', are

respectively the semi-perimeters of the triangles. Draw B'F perpendi-

cular to BC, and cut off AH equal to AB, and join BH. If BF can be

proved to be greater than B'H, the perimeter of ABC is greater than the

perimeter of AB'C. Next let B C fall above BC.

10. The angles contained in the two segments of the circle, may be

shewn to be equal, then by joining the extremities of the arcs, the two

remaining sides may be shewn to be parallel.

11. It may be shewn that four equal and equilateral triangles will

form an equilateral triangle of the same perimeter as the hexagon, which

is formed by six equal and equilateral triangles.

12. Let the figure be constructed. By drawing the diagonals of the

hexagon, the proof is obvious.

13. By Euc. I. 47, the perpendicular distance from the center of the

circle upon the side of the inscribed hexagon may be found.

14. The alternate sides of the hexagon will fall upon the sides of the

:riangle, and each side will be found to be equal to one-third of the side

Df the equilateral triangle.

15. A regular duodecagon may be inscribed in a circle by means of

the equilateral triangle and square, or by means of the hexagon. The

irea of the duodecagon is three times the square on the radius of the circle,

which is the square on the side of an equilateral triangle inscribed in the

same circle. Theorem 1, p. 196.

16. In general, three straight lines when produced will meet and

hrm. a triangle, except when all three are parallel or two parallel are

ntersected by the third. This Problem includes Euc. iv. 5, and all the

ases which arise from producing the sides of the triangle. The circles

lescribed touching a side of a triangle and the other two sides produced,

ire called the escribed circles.

17. This is manifest from Euc. iii. 21.

18. The point required is the center of the circle which circumscribes

;he triangle. See the notes on Euc. iii. 20, p. 155.

19. if the perpendiculars meet the three sides of the triangle, the

)oint is within the triangle, Euc. iv. 4. If the perpendiculars meet the

)ase and the two sides produced, the point is the center of the escribed

ircle.

20. This is manifest from Euc. in. 11, 18.

21. The base BC is intersected by the perpendicular AD, and the

ide AC is intersected by the perpendicular BE. From Theorem i. p.

60 ; the arc AF is proved equal to AE, or the arc FE is bisected in A.

!n the same manner the arcs FD, DE, may be shewn to be bisected inBC.

22. Let ABC be a triangle, and let D, E be the points where the in-

cribed circle touches the sides AB, AC. Draw BE, CD intersecting

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