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parallels.

6. A straight line drawn between two parallels and terminated by

them, is bisected ; shew that any other straight line passing through

the middle point and terminated by the parallels, is also bisected at

that point.

7. If through a point equidistant from two parallel straight lines,

two straight lines are drawn cutting the parallels, the portions of the

latter thus intercepted are equal.

Pboblems.

8. AB and CD are two given straight lines, and X is a given

point in AB : find a point Y in AB such that YX may be equal to the

perpendicular distance of Y from CD.

9. ABC is an isosceles triangle; required to draw a straight

line DE parallel to the base BC, and meeting the equal sides in D and

E, so that BD, DE, EC may be all equal.

10. ABC is any triangle; required to draw a straight line DE

parallel to the base BC, and meeting the other sides in D and E, so

that DE may be equal to the sum of BD and CE.

11. ABC is any triangle; required to draw a straight line parallel

to the base BC, and meeting the other sides in D and E, so that DE

may be equal to the difference of BD and CE.

96

EUCLIDS ELEMENTS.

IV. ON PARALLELOGRAMS.

See Propositions 33, 34, and the deductions from these Props,

jiven on page 64.

1. The straight line drawn through the middle point of a side of a

triangle parallel to the base, bisects the remaining side.

Let ABC be a A , and Z the middle point

of the side AB. Through Z, Z Y is drawn par'

to BC ; then shall Y be the middle point of AC.

Through Z draw ZX par^ to AC. i.31.

Then in the a^ AZY, ZBX,

because ZY and BC are par',

.-.the Z AZY=:the z ZBX; i. 29.

and because ZX and AC are par',

.-. the Z ZAY = the z BZX; i. 29.

alsoAZ = ZB: Hijj>.

.-. AY = ZX.

But ZXCY is a parÂ»i by construction ;

.-. ZX=:YC.

Hence AY = YC;

that is, AC is bisected at Y. q.e.ij.

2. The straight line 7vhich joins the middle points of two sides of a

triangle, is parallel to the third side.

Let ABC be a A , and Z, Y the middle

points of the sides AB, AC:

then shall ZY be par' to BC.

Produce ZY to V, making YV equal to

ZY.

Join CV.

Then in the a^AYZ.CYV,

( AY = CY, Hyp.

'' andYZ=YV, Constr.

Because

YV,

I and the Z AYZ = the vert, opp

.-. AZ = CV,

and the Z ZAY = the Z VCY;

hence CV is par' to AZ. i. 27.

But CV is equal to AZ, that is, to BZ : Hgp.

.'. CV is equal and par' to BZ :

.-. ZV is equal and par' to BC : i. 33.

that is, ZY is par' to BC. q.e.d.

[A second proof of this proposition may be derived from i. 38, 39.]

THEOREMS AND EXAMPLES ON BOOK I. 97

3. The straight line which joins the middle points of two sides of a

triangle is equal to half the third side.

4. Shew that the three straight lines which join the middle points

of the sides of a triangle, divide it into four triangles which are identi-

cally equal.

5. Any straight line draion from the vertex of a triangle to the

base is bisected by the straight line ichich joins the middle points of the

other sides of the triangle.

6. Given the three middle points of the sides of a triangle, con-

struct the triangle.

7. AB, AC are two given straight lines, and P is a given point

between them ; required to draw through P a straight line termi-

nated by AB, AC, and bisected by P.

8. ABCD is a parallelogram, and X, Y are the middle points of

the opposite sides AD, BC: shew that BX and DY trisect the dia-

gonal AC.

9. If the middle points of adjacent sides of any quadrilateral he

joined, thejigure thus formed is a parallelogram.

10. Shew that the straight lines which join the middle points of

opposite sides of a quadrilateral, bisect one another.

11. The straight line which joins the middle points of the oblique

sides of a trapezium, is parallel to the two parallel sides, and passes

through the middle points of the diagonals.

12. TJie straight line which joins the middle points of the oblique

sides of a trapezium is equal to half the sum of the parallel sides ; and

the portion intercepted between the diagonals is equal to half the

difference of the parallel sides.

Definition. If from the extremities of one straight line per-

pendiculars are drawn to another, the portion of the latter

intercepted between the perpendiculars is said to be the Ortho-

goiial Projection of the tirst line upon the second.

B

B

Y Q

Thus in the adjoining figures, if from the extremities of the straight

hue AB the perpendiculars AX, BY are drawn to PQ, then XY is the

orthogonal projection of AB on PQ.

H. E.

98

EUCLID'S ELEMENTS.

13. A given straight line AB is bisected at C; shew that the i)ro-

jections o/AC, CB oÂ» any other straight line are equal.

c

X ^

E

z

\

Y

p ^

Q

Let XZ, Z Y be the projections of AC, CB on any straight line 90. :

then XZ and ZY shall be equal.

Through A draw a straight line parallel to PQ, meeting CZ, BY

or these lines produced, in H, K. i. 31.

Now AX, CZ, BY are parallel, for they are perp. to PQ; i. 28.

.-. the figures XH, HY are parÂ°";

.-. AH-XZ, and HK = ZY. i. 34.

But through C, the middle point of AB, a side of the a ABK,

CH has been drawn parallel to the side BK ;

.-. CH bisects AK: Ex. 1, p. 96.

that is, AH = HK;

.-. XZ=ZY. Q.E.D.

14. 1/ three parallel straight lines make equal intercepts on a

fourth straight line which meets them, they will also make equal inter-

cepts on any other straight line 2ohich meets them.

15. Equal and parallel straight lines have equal projections on any

other straight line.

16. AB is a given straight line bisected at O ; and AX, BY are

perpendiculars drawn from A and B on any other straight line : shew

that OX is equal to OY.

17. AB is a given straight line bisected at O : and AX, BY and OZ

are perpendiculars dratvn to any straight line PQ, lohich does not pass

between A and B: shew that OZ is equal to half the sum of AX, BY.

[OZ is said to be the Aritlunetie Mean between AX and BY.]

18. AB is a given straight line bisected at O; and through A, B

and O parallel straight lines are drawn to meet a given straight line

PQ in X, Y, Z : shew that OZ is equal to half the sum, or half the

difference of AX and BY, according as A and B lie on the same side

or on opposite sides of PQ.

THEOREMS AND EXAMPLES ON BOOK I. 99

A

19. To divide a given Jinite straight line into any nuinher of equal

parts.

[For example, required to divide the straight

line AB into Jive equal parts.

From A draw AC, a straight line of un-

limited length, making any angle with AB.

In AC take any point P, and mark off

successive parts PQ, QR, RS, ST each equal

to A P.

Join BT ; and through P, Q, R, S draw

parallels to BT.

It may be shewn by Ex. 14, p. 98, that these

parallels divide AB into five equal parts.]

20. If through an angle of a parallelogram any straight line

is draicn, the perpendicular drawn to it from the opposite angle

is equal to the sum or difference of the perpendiculars drawn to it

from the two remaining angles, according as the given straight line

falls without the parallelogram, or intersects it.

[Through the opposite angle draw a straight line parallel to the

given straight line, so as to meet the perpendicular from one of the

remaining angles, produced if necessary: then apply i. 34, i. 26. Or

proceed as in the following example.]

21. From the angular points of a parallelogram perpendiculars

are drawn to any straight line which is without the parallelogram:

shew that the sum of the perpendiculars drawn from one pair of

opposite angles is equal to the sum of those drawn from the other pair.

[Draw the diagonals, and from their point of intersection let fall a

perpendicular upon the given straight line. See Ex. 17, p. 98.]

22. The sum of the perpendiculars drawn from any point in the

base of an isosceles triangle to the equal sides is equal to the perpendi-

cular drawn from either extremity of the base to the opposite side.

[It follows that the sum of the distances of any point in the base

of an isosceles triangle from the equal sides is constant, that is,

the same whatever point in the base is taken.]

23. In the base produced of an isosceles triangle any point is

taken : shew that the difference of its distances from the equal sides is

constant.

24. The sum of the perpendiculars drawn from any point within

an equilateral triangle to the three sides is equal to the perpendicular

drawn from any one of the angular points to the opposite side, and is

therefore constant.

7â€”2

100 Euclid's elements,

PROBLEMB.

[Problems marked (*) admit of more than one solution.]

*25. Draw a straight line through a given point, so that the part of

it intercepted between two given parallel straight lines may be of given

length.

26. Draw a straight line parallel to a given straight line, so that

the part intercepted between two other given straight lines may be of

given length.

27. Draw a straight line equally inclined to two given straight

lines that meet, so that the part intercepted between them may be of

given length.

28. AB, AC are two given straight lines, and P is a given point

loithout the angle contained by them. It is required to draw through

P a straight line to meet the given lines, so that the part intercepted

between them may be equal to the part between P and the nearer line.

v. MISCELLANEOUS THEOREMS AND EXAMPLES.

Chiefly on i. 32.

1. A is the vertex of an isosceles triangle ABC, and BA is jjroduced

to D, so that AD is equal to BA ; if DC is draicn, sheio that BCD is a

right angle.

2. The straight line joining the middle point of the hypotenuse of a

right-angled triangle to the right angle is equal to half the hypotenuse.

3. From the extremities of the base of a triangle perpendiculars

are drawn to the opposite sides (produced if necessary) ; shew that the

straight lines which join the middle point of the base to the. feet of

the perpendiculars are equal.

4. In a triangle ABC, AD is draxcn perpendicular to BC ; and

X, Y, Z are the middle points of the sides BC, CA, AB respectively :

shexo that each of the angles ZXY, ZDY is equal to the angle BAC.

5. In a right-angled triangle, if a j)erpendicular he drawn from

the right angle to the hypotenuse, the two triangles thus formed are

equiangular to one another.

6. In a right-angled triangle tico straight lines are draion from,

the right angle, one bisecting the hypotenuse, the other perpendicular

to it : shew that they contain an angle equal to the difference of the two

acute angles of the irlaiigle. [See above, Ex. 2 and Ex. 5.]

THEOREMS AND EXAMPLES ON BOOK I. 101

7. In a triangle if a 'perpendicular he draicn from one extremity

of the base to the bisector of the vertical angle, (i) it ivill make with

either of the sides containing the vertical angle an angle equal to half

the sum of the angles at the base; (ii) it will make with the base an

angle equal to half the difference of the angles at the base.

Let ABC be the given a, and AH the bi-

sector of the vertical z BAG,

Let CLK meet AH at right angles.

(i) Then shall each of the Z ^ AKC, ACK

be equal to half the sum of the z " ABC,

ACB.

In the A^AKL, ACL,

r the z KAL = the z CAL,

Because < also the Z ALK = the Z ALC, being rt.

( and AL is common to both a";

.-. the Z AKL = the Z ACL. i. 26.

Again, the z AKC = the sum of the z ' KBC, KCB ; i. 32.

that is, the z ACK = the sum of the z " KBC, KCB.

To each add the Z ACK,

then twice the Z ACK = the sum of the z Â« ABC, ACB,

.-. the Z ACK=half thesumof the z Â« ABC, ACB.

(ii) The z KCB shall be equal to half the difference of the

Z^ACB, ABC.

As before, the z ACK = the sum of the Z Â« KBC, KCB,

To each of these add the z KCB :

then the Z ACB = the Z KBC together with twice the z KCB.

.-. twice the Z KCB = the difference of the Z^ACB, KBC,

that is, the z KCB=half the difference of the Z ^ ACB, ABC.

CoKOLLAKY. If X be the middle point of the base, and XL be joined,

it may be shewn by Ex. 3, p. 97, that XL is half BK ; that is, that

XL is half the difference of the sides AB, AC.

8. In any triangle the angle contained by the bisector of the

vertical angle and the perpendicular from the vertex to the base is equal

to half the difference of the angles at the base. [See Ex. 3, p. 59.]

9. In a triangle ABC the side AC is produced to D, and the

angles BAC, BCD are bisected by straight lines which meet at F;

shew that they contain an angle equal to half the angle at B.

10. If in a right-angled triangle one of the acute angles is double

of the other, shew that the hypotenuse is double of the shorter side.

11. If in a diagonal of a parallelogram any two points equidistant

from its extremities be joined to the opposite angles, the figure thus

formed will be also a parallelogram.

T02 KUCLIirt^ ELEMENTS.

12. ABC is a given equilateral triangle, and in the sides BC, CA,

AB the points X, Y, Z are taken respective!}', so that BX, CY and AZ

are all equal. AX, BY, CZ are now drawn, intersecting in P, Q, R :

shew that the triangle PQR is equilateral.

13. If in the sides AB, BC, CD, DA of a parallelogram ABCD

four points P, Q, R, S be taken in order, one in each side, so that AP,

BQ, CR, DS are all equal; shew that the figure PQRS is a parallelo-

gram. â€¢

14. In the figure of i. 1, if the circles intersect at F, and if

CA and CB are produced to meet the circles in P and Q. respectively ;

shew that the points P, F, Q are in the same straight line; and

shew also that the triangle CPQ is equilateral.

[Problems marked (*) admit of more than one solution.]

15. Through two given points draw two straight lines forming

with a straight line given in position, an equilateral triangle.

*16. From a given point it is required to draw to two parallel

straight lines two equal straight lines at right angles to one another.

*17. Three given straight lines meet at a point ; draw another

straight line so that the two portions of it intercepted between the

given lines may be equal to one another.

18. From a given point draw three straight lines of given lengths,

so that their extremities may be in the same straight line, and inter-

cept equal distances on that line. [See Fig. to i. 16.]

19. Use the properties of the equilateral triangle to trisect a given

finite straight line.

20. In a given triangle inscribe a rhombus, having one of its

angles coinciding with an angle of the triangle.

VI. ON THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.

Definitions, (i) Three or more straight lines are said to

l>e concurrent when they meet in one point.

(ii) Three or more points are said to be collinear when they

lie upon one straight line.

We here give some propositions relating to the concurrence

of certain groups of straight lines drawn in a triangle : the im-

portance of these theorems will be more fully appreciated when

the student is familiar with Books iii. and iv.

THEOREMS AND EXAMPLES ON BOOK I.

103

1. The perpendiculars drawn to the sides of a triangle from their

middle points are concurrent.

Let ABC be a A, and X, Y, Z the

middle points of its sides :

then shall the perp^ drawn to the

sides from X, Y, Z be concurrent.

From Z and Y draw perps to A B, AC;

these perps, since they cannot be parallel,

will meet at point O. Ax. 12.

Join OX.

Because

Tt is required to prove that OX is perp. to BC

Join OA, OB, OC.

Inthe AÂ« OYA, OYC,

YA = YC,

and OY is common to both ;

I also the z OYA = the Z OYC, being rt.L\

.-. OA = OC.

Similarly, from the a^ OZA, OZB,

it may be proved that OA = OB.

Hence OA, OB, OC are all equal.

Again, in the aÂ« OXB, OXC

( BX = CX,

Because <and XO is common to both ;

( also OB = 00:

.-. the Z OXB = the Z OXC;

but these are adjacent z ^ ;

.-. they are rt. l^;

that is, OX is perp. to BC.

Hence the three perps OX, OY, OZ meet in the point O.

Q. E

2. The bisectors of the angles of a triangle are concurrent.

Let ABC be a A . Bisect the Z Â« ABC, A

BOA, by straight lines which must meet

at some point O. Ax. 12.

Join AO.

It is required to prove that AO bisects the

Z BAG.

From O draw OP, OQ, OR perp. to the

sides of the a .

Then in the a** OBP, OBR,

I the Z OBP = the Z OBR,

Because .| and the z OPB = the z ORB, being rt. L^,

( and OB is common ;

.-. OP = OR.

Hy2).

1.4.

Hyp-

Proved.

1.8.

Def. 7.

I. 20.

104

EUCLID'S ELEMENTS.

Similarly from the a" OCP, OCQ,

it may be shewn that OP=OQ,

.-. OP, OQ, OR are all equal.

Again in the a" ORA, OQA,

I the z" ORA, OQA are rt. l',

P land the hypotenuse OA is

liecause. common,

[ also OR -OQ; Proved.

.'. the z RAO = the z QAO. Ex. 12, p. 91.

That is, AO is the bisector of the z BAG.

Hence the bisectors of the three z ** meet at the point O.

Q. E. r>.

3. The bisectors of two exterior angles of a triangle and tlie

bisector of the third angle are concurrent.

Let ABC be a a, of which the sides AB,

AC are produced to any points D and E.

Bisect the Z^ DBC, ECB by straight lines

which must meet at some point O. Ax. 12.

Join AO.

It is required to inove that AO bisects the

angle BAC.

From O draw OP, OQ, OR perp. to the

sides of the a .

Then in the aÂ« OBP, OBR,

the Z OBP = the z OBR, Gomtr.

also the z OPB = the z ORB,

being rt. l ",

and OB is common ;

.-. OP = OR.

Similarly in the a" OCP, OCQ,

it may be shewn that OP = OQ:

.-. OP, OQ, OR are all equal.

Again in the A' ORA, OQA,

( the Z " ORA, OQA are rt. l %

Because \ and the hypotenuse OA is common,

( alsoOR = OQ;

.-. the z RAO = the Z QAO.

Because

Proved.

Ex. 12, p. 91.

That is, AO is the bisector of the z BAC.

.-. the bisectors of the two exterior Z " DBC, ECB,

and of tlie interior z BAC meet at the point O.

Q.B.D.

THEOREMS AND EXAMPLES ON BOOK I.

105

4. The medians of a triangle are concurrent.

Let ABC be a a . Let BY and CZ be two of its

medians, and let them intersect at O.

Join AO,

and produce it to meet BC in X.

It is required to shew that AX is the remaining

median of the a .

Through C draw CK parallel to BY:

produce AX to meet CK at K.

Join BK.

In the A AKC,

because Y is the middle point of AC, and- YD is

parallel to CK,

. â€¢. O is the middle point of AK. Ex. 1, p. 96.

Again in the a ABK,

since Z and O are the middle points of AB, AK,

.-. ZO is parallel to BK, Ex. 2, p. 96.

that is, OC is parallel to BK :

.-. the figure BKCO is a par".

But the diagonals of a parâ„¢ bisect one another, Ex. 5, p. 64.

.-. X is the middle point of BC.

That is, AX is a median of the a .

Hence the three medians meet at the point O. q.e.d.

Corollary. The three medians of a triangle cut one another at a

point of trisection, the greater segment in each being toioards the

angular point.

For in the above figure it has been proved that

AO = OK,

also that OX is half of OK;

.-. OX is half of OA :

that is, OX is one third of AX.

Similarly OY is one third of BY,

and OZ is one third of CZ. q.e.d.

By means of this Corollary it may be shewn that in any triangle

the shorter median bisects the greater side.

[The point of intersection of the three medians of a triangle is

called the centroid. It is shewn in mechanics that a thin triangular

])late will balance in any position about this point : therefore the

centroid of a triangle is also its centre of gravity.]

106

EUCLIDS ELEMENTS.

5. 21ie perpendiculars drawn from the verticeit of a triaiiaJ<' to tJir

opposite sides are concurrent.

Let ABC be a A, and AD, BE, CF the three perp" drawn from

the vertices to the opposite sides :

then shall these perp* be concurrent.

Through A, B, and C draw straight lines MN, NL, LM parallel

to the opposite sides of the a .

Then the figure BAMC is a par'". Bef. 26.

.-. AB=MC. 1.34.

Also the figure BACL is a par".

.-. AB=LC,

.-. LC = CM :

that is, C is the middle point of LM.

So also A and B are the middle points of M N and N L.

Hence AD, BE, CF are the perp* to the sides of the a LMN from

their middle points. Ex. 3, p. 54.

But these perp' meet in a point: Ex. 1, p. 103.

that is, the perp* drawn from the vertices of the a ABC to the

opposite sides meet in a point. q.e.d.

[For another proof see Theorems and Examples on Book iii.]

Definitions.

(i; The intersection of the perpendiculars drawn from the

vertices of a triangle to the oi)posite sides is called its ortho-

centre.

(ii) The triangle formed b}^ joining the feet of the perpen-

diculars is called the pedal triangle.

THEOREMS AND EXAMPLES ON BOOK I.

107

K

VII. ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS.

No general rules can be laid down for the solution of

problems in this section; but in a few typical cases we give

constructions, which the student will find little difficulty in

adapting to other questions of the same class.

1. Construct a right-angled triangle, having given the hypotenuse

and the sum of the remaining sides.

[It is required to construct a rt. ~

angled A , having its hypotenuse equal

to the given straight line K, and the sum

of its remaining sides equal to AB.

From A draw AE making with BA

an z equal to half a rt. l . From

centre B, with radius equal to K, de-

scribe a circle cutting AE in the points

cc.

From C and C draw perp' CD, CD' to AB; and join CB, C'B.

Then either of the a^ CDB, C'D'B will satisfy the given conditions.

Note, If the given hypotenuse K be greater than the perpendicu-

lar drawn from B to AE, there will be two solutions. If the line K be

equal to this perpendicular, there will be one solution ; but if less, the

problem is impossible.]

2. Construct a right-angled triangle, having given the hypotenuse

and the difference of the remaining sides.

3. Construct an isosceles right-angled triangle, having given the

sum of the hypotenuse and one side.

4. Construct a triangle, having given the perimeter and the angles

at the base.

R B

J

[Let AB be the perimeter of the required a , and X and Y the

the base.

From A draw AP, making the / BAP equal to half the z X.

From B draw BP, making the Z ABP equal to half the z Y,

From P draw PQ, making the Z APQ equal to the z BAP.

From P draw PR, making the z BPR equal to the z ABP.

Then shall PQR be the required a .]

Z%at

108 EUCLID'S ELEMENTS.

5. Coustruct a right-angled triangle, having given tlie perimeter

and one acute angle.

6. Construct an isosceles triangle of given altitude, so that its

base may be in a given straight line, and its two equal sides may pass

through two fixed points. [See Ex. 7, p. 49.]

7. Construct an equilateral triangle, having given the length of

the perpendicular drawn from one of the vertices to the opposite side.

*- 8. Construct an isosceles triangle, having given the base, and

the difference of one of the remaining sides and the perpendicular

drawn from the vertex to the base. [See Ex. 1, p. 88.]

9. Construct a triangle, having given the base, one of the angles

at the base, and the sum of the remaining sides.

10. Construct a triangle, having given the base, one of the angles

at the base, and the difference of the remaining sides. ,

Nil. Construct a triangle, liaving given the base, the difference

of the angles at the base, and tJie difference of the remaining sides.

k

[Let AB be the given base, X the difference of the z ' at the base,

and K the difference of the remaining sides.

Draw BE, making the Z ABE equal to half the Z X.

From centre A, with radius equal to K, describe a circle cutting BE

in D and D'. Let D be the point of intersection nearer to B.

Join AD and produce it to C.

Draw BC, making the z DBC equal to the z BDC.

Then shall CAB be the a required. Ex. 7, p. 101.

Note. This problem is possible only when the given difference K

is greater than the perpendicular drawn from A to BE.]

~12. Construct a triangle, having given the base, the difference of

the angles at the base, and the sum of the remaining sides.

"HES. Construct a triangle, having given the perpendicular from the

vertex on the base, and the difference between each side and the

adjacent segment of the base.

THEOREMS AND EXAMPLES ON BOOK I. 109

14. Construct a triangle, having given two sides and the median

which bisects the remaining side, [See Ex. 18, p. 102.]

15. Construct a triangle, having given one side, and the medians

which bisect the two remaining sides.

[See Fig. to Ex. 4, p. 105.

Let BC be the given side. Take two-thirds of each of the given

medians; hence construct the triangle BOC. The rest of the con-

struction follows easily.]

16. Construct a triangle, having given its three medians.

[See Fig. to Ex. 4, p. 105.

6. A straight line drawn between two parallels and terminated by

them, is bisected ; shew that any other straight line passing through

the middle point and terminated by the parallels, is also bisected at

that point.

7. If through a point equidistant from two parallel straight lines,

two straight lines are drawn cutting the parallels, the portions of the

latter thus intercepted are equal.

Pboblems.

8. AB and CD are two given straight lines, and X is a given

point in AB : find a point Y in AB such that YX may be equal to the

perpendicular distance of Y from CD.

9. ABC is an isosceles triangle; required to draw a straight

line DE parallel to the base BC, and meeting the equal sides in D and

E, so that BD, DE, EC may be all equal.

10. ABC is any triangle; required to draw a straight line DE

parallel to the base BC, and meeting the other sides in D and E, so

that DE may be equal to the sum of BD and CE.

11. ABC is any triangle; required to draw a straight line parallel

to the base BC, and meeting the other sides in D and E, so that DE

may be equal to the difference of BD and CE.

96

EUCLIDS ELEMENTS.

IV. ON PARALLELOGRAMS.

See Propositions 33, 34, and the deductions from these Props,

jiven on page 64.

1. The straight line drawn through the middle point of a side of a

triangle parallel to the base, bisects the remaining side.

Let ABC be a A , and Z the middle point

of the side AB. Through Z, Z Y is drawn par'

to BC ; then shall Y be the middle point of AC.

Through Z draw ZX par^ to AC. i.31.

Then in the a^ AZY, ZBX,

because ZY and BC are par',

.-.the Z AZY=:the z ZBX; i. 29.

and because ZX and AC are par',

.-. the Z ZAY = the z BZX; i. 29.

alsoAZ = ZB: Hijj>.

.-. AY = ZX.

But ZXCY is a parÂ»i by construction ;

.-. ZX=:YC.

Hence AY = YC;

that is, AC is bisected at Y. q.e.ij.

2. The straight line 7vhich joins the middle points of two sides of a

triangle, is parallel to the third side.

Let ABC be a A , and Z, Y the middle

points of the sides AB, AC:

then shall ZY be par' to BC.

Produce ZY to V, making YV equal to

ZY.

Join CV.

Then in the a^AYZ.CYV,

( AY = CY, Hyp.

'' andYZ=YV, Constr.

Because

YV,

I and the Z AYZ = the vert, opp

.-. AZ = CV,

and the Z ZAY = the Z VCY;

hence CV is par' to AZ. i. 27.

But CV is equal to AZ, that is, to BZ : Hgp.

.'. CV is equal and par' to BZ :

.-. ZV is equal and par' to BC : i. 33.

that is, ZY is par' to BC. q.e.d.

[A second proof of this proposition may be derived from i. 38, 39.]

THEOREMS AND EXAMPLES ON BOOK I. 97

3. The straight line which joins the middle points of two sides of a

triangle is equal to half the third side.

4. Shew that the three straight lines which join the middle points

of the sides of a triangle, divide it into four triangles which are identi-

cally equal.

5. Any straight line draion from the vertex of a triangle to the

base is bisected by the straight line ichich joins the middle points of the

other sides of the triangle.

6. Given the three middle points of the sides of a triangle, con-

struct the triangle.

7. AB, AC are two given straight lines, and P is a given point

between them ; required to draw through P a straight line termi-

nated by AB, AC, and bisected by P.

8. ABCD is a parallelogram, and X, Y are the middle points of

the opposite sides AD, BC: shew that BX and DY trisect the dia-

gonal AC.

9. If the middle points of adjacent sides of any quadrilateral he

joined, thejigure thus formed is a parallelogram.

10. Shew that the straight lines which join the middle points of

opposite sides of a quadrilateral, bisect one another.

11. The straight line which joins the middle points of the oblique

sides of a trapezium, is parallel to the two parallel sides, and passes

through the middle points of the diagonals.

12. TJie straight line which joins the middle points of the oblique

sides of a trapezium is equal to half the sum of the parallel sides ; and

the portion intercepted between the diagonals is equal to half the

difference of the parallel sides.

Definition. If from the extremities of one straight line per-

pendiculars are drawn to another, the portion of the latter

intercepted between the perpendiculars is said to be the Ortho-

goiial Projection of the tirst line upon the second.

B

B

Y Q

Thus in the adjoining figures, if from the extremities of the straight

hue AB the perpendiculars AX, BY are drawn to PQ, then XY is the

orthogonal projection of AB on PQ.

H. E.

98

EUCLID'S ELEMENTS.

13. A given straight line AB is bisected at C; shew that the i)ro-

jections o/AC, CB oÂ» any other straight line are equal.

c

X ^

E

z

\

Y

p ^

Q

Let XZ, Z Y be the projections of AC, CB on any straight line 90. :

then XZ and ZY shall be equal.

Through A draw a straight line parallel to PQ, meeting CZ, BY

or these lines produced, in H, K. i. 31.

Now AX, CZ, BY are parallel, for they are perp. to PQ; i. 28.

.-. the figures XH, HY are parÂ°";

.-. AH-XZ, and HK = ZY. i. 34.

But through C, the middle point of AB, a side of the a ABK,

CH has been drawn parallel to the side BK ;

.-. CH bisects AK: Ex. 1, p. 96.

that is, AH = HK;

.-. XZ=ZY. Q.E.D.

14. 1/ three parallel straight lines make equal intercepts on a

fourth straight line which meets them, they will also make equal inter-

cepts on any other straight line 2ohich meets them.

15. Equal and parallel straight lines have equal projections on any

other straight line.

16. AB is a given straight line bisected at O ; and AX, BY are

perpendiculars drawn from A and B on any other straight line : shew

that OX is equal to OY.

17. AB is a given straight line bisected at O : and AX, BY and OZ

are perpendiculars dratvn to any straight line PQ, lohich does not pass

between A and B: shew that OZ is equal to half the sum of AX, BY.

[OZ is said to be the Aritlunetie Mean between AX and BY.]

18. AB is a given straight line bisected at O; and through A, B

and O parallel straight lines are drawn to meet a given straight line

PQ in X, Y, Z : shew that OZ is equal to half the sum, or half the

difference of AX and BY, according as A and B lie on the same side

or on opposite sides of PQ.

THEOREMS AND EXAMPLES ON BOOK I. 99

A

19. To divide a given Jinite straight line into any nuinher of equal

parts.

[For example, required to divide the straight

line AB into Jive equal parts.

From A draw AC, a straight line of un-

limited length, making any angle with AB.

In AC take any point P, and mark off

successive parts PQ, QR, RS, ST each equal

to A P.

Join BT ; and through P, Q, R, S draw

parallels to BT.

It may be shewn by Ex. 14, p. 98, that these

parallels divide AB into five equal parts.]

20. If through an angle of a parallelogram any straight line

is draicn, the perpendicular drawn to it from the opposite angle

is equal to the sum or difference of the perpendiculars drawn to it

from the two remaining angles, according as the given straight line

falls without the parallelogram, or intersects it.

[Through the opposite angle draw a straight line parallel to the

given straight line, so as to meet the perpendicular from one of the

remaining angles, produced if necessary: then apply i. 34, i. 26. Or

proceed as in the following example.]

21. From the angular points of a parallelogram perpendiculars

are drawn to any straight line which is without the parallelogram:

shew that the sum of the perpendiculars drawn from one pair of

opposite angles is equal to the sum of those drawn from the other pair.

[Draw the diagonals, and from their point of intersection let fall a

perpendicular upon the given straight line. See Ex. 17, p. 98.]

22. The sum of the perpendiculars drawn from any point in the

base of an isosceles triangle to the equal sides is equal to the perpendi-

cular drawn from either extremity of the base to the opposite side.

[It follows that the sum of the distances of any point in the base

of an isosceles triangle from the equal sides is constant, that is,

the same whatever point in the base is taken.]

23. In the base produced of an isosceles triangle any point is

taken : shew that the difference of its distances from the equal sides is

constant.

24. The sum of the perpendiculars drawn from any point within

an equilateral triangle to the three sides is equal to the perpendicular

drawn from any one of the angular points to the opposite side, and is

therefore constant.

7â€”2

100 Euclid's elements,

PROBLEMB.

[Problems marked (*) admit of more than one solution.]

*25. Draw a straight line through a given point, so that the part of

it intercepted between two given parallel straight lines may be of given

length.

26. Draw a straight line parallel to a given straight line, so that

the part intercepted between two other given straight lines may be of

given length.

27. Draw a straight line equally inclined to two given straight

lines that meet, so that the part intercepted between them may be of

given length.

28. AB, AC are two given straight lines, and P is a given point

loithout the angle contained by them. It is required to draw through

P a straight line to meet the given lines, so that the part intercepted

between them may be equal to the part between P and the nearer line.

v. MISCELLANEOUS THEOREMS AND EXAMPLES.

Chiefly on i. 32.

1. A is the vertex of an isosceles triangle ABC, and BA is jjroduced

to D, so that AD is equal to BA ; if DC is draicn, sheio that BCD is a

right angle.

2. The straight line joining the middle point of the hypotenuse of a

right-angled triangle to the right angle is equal to half the hypotenuse.

3. From the extremities of the base of a triangle perpendiculars

are drawn to the opposite sides (produced if necessary) ; shew that the

straight lines which join the middle point of the base to the. feet of

the perpendiculars are equal.

4. In a triangle ABC, AD is draxcn perpendicular to BC ; and

X, Y, Z are the middle points of the sides BC, CA, AB respectively :

shexo that each of the angles ZXY, ZDY is equal to the angle BAC.

5. In a right-angled triangle, if a j)erpendicular he drawn from

the right angle to the hypotenuse, the two triangles thus formed are

equiangular to one another.

6. In a right-angled triangle tico straight lines are draion from,

the right angle, one bisecting the hypotenuse, the other perpendicular

to it : shew that they contain an angle equal to the difference of the two

acute angles of the irlaiigle. [See above, Ex. 2 and Ex. 5.]

THEOREMS AND EXAMPLES ON BOOK I. 101

7. In a triangle if a 'perpendicular he draicn from one extremity

of the base to the bisector of the vertical angle, (i) it ivill make with

either of the sides containing the vertical angle an angle equal to half

the sum of the angles at the base; (ii) it will make with the base an

angle equal to half the difference of the angles at the base.

Let ABC be the given a, and AH the bi-

sector of the vertical z BAG,

Let CLK meet AH at right angles.

(i) Then shall each of the Z ^ AKC, ACK

be equal to half the sum of the z " ABC,

ACB.

In the A^AKL, ACL,

r the z KAL = the z CAL,

Because < also the Z ALK = the Z ALC, being rt.

( and AL is common to both a";

.-. the Z AKL = the Z ACL. i. 26.

Again, the z AKC = the sum of the z ' KBC, KCB ; i. 32.

that is, the z ACK = the sum of the z " KBC, KCB.

To each add the Z ACK,

then twice the Z ACK = the sum of the z Â« ABC, ACB,

.-. the Z ACK=half thesumof the z Â« ABC, ACB.

(ii) The z KCB shall be equal to half the difference of the

Z^ACB, ABC.

As before, the z ACK = the sum of the Z Â« KBC, KCB,

To each of these add the z KCB :

then the Z ACB = the Z KBC together with twice the z KCB.

.-. twice the Z KCB = the difference of the Z^ACB, KBC,

that is, the z KCB=half the difference of the Z ^ ACB, ABC.

CoKOLLAKY. If X be the middle point of the base, and XL be joined,

it may be shewn by Ex. 3, p. 97, that XL is half BK ; that is, that

XL is half the difference of the sides AB, AC.

8. In any triangle the angle contained by the bisector of the

vertical angle and the perpendicular from the vertex to the base is equal

to half the difference of the angles at the base. [See Ex. 3, p. 59.]

9. In a triangle ABC the side AC is produced to D, and the

angles BAC, BCD are bisected by straight lines which meet at F;

shew that they contain an angle equal to half the angle at B.

10. If in a right-angled triangle one of the acute angles is double

of the other, shew that the hypotenuse is double of the shorter side.

11. If in a diagonal of a parallelogram any two points equidistant

from its extremities be joined to the opposite angles, the figure thus

formed will be also a parallelogram.

T02 KUCLIirt^ ELEMENTS.

12. ABC is a given equilateral triangle, and in the sides BC, CA,

AB the points X, Y, Z are taken respective!}', so that BX, CY and AZ

are all equal. AX, BY, CZ are now drawn, intersecting in P, Q, R :

shew that the triangle PQR is equilateral.

13. If in the sides AB, BC, CD, DA of a parallelogram ABCD

four points P, Q, R, S be taken in order, one in each side, so that AP,

BQ, CR, DS are all equal; shew that the figure PQRS is a parallelo-

gram. â€¢

14. In the figure of i. 1, if the circles intersect at F, and if

CA and CB are produced to meet the circles in P and Q. respectively ;

shew that the points P, F, Q are in the same straight line; and

shew also that the triangle CPQ is equilateral.

[Problems marked (*) admit of more than one solution.]

15. Through two given points draw two straight lines forming

with a straight line given in position, an equilateral triangle.

*16. From a given point it is required to draw to two parallel

straight lines two equal straight lines at right angles to one another.

*17. Three given straight lines meet at a point ; draw another

straight line so that the two portions of it intercepted between the

given lines may be equal to one another.

18. From a given point draw three straight lines of given lengths,

so that their extremities may be in the same straight line, and inter-

cept equal distances on that line. [See Fig. to i. 16.]

19. Use the properties of the equilateral triangle to trisect a given

finite straight line.

20. In a given triangle inscribe a rhombus, having one of its

angles coinciding with an angle of the triangle.

VI. ON THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.

Definitions, (i) Three or more straight lines are said to

l>e concurrent when they meet in one point.

(ii) Three or more points are said to be collinear when they

lie upon one straight line.

We here give some propositions relating to the concurrence

of certain groups of straight lines drawn in a triangle : the im-

portance of these theorems will be more fully appreciated when

the student is familiar with Books iii. and iv.

THEOREMS AND EXAMPLES ON BOOK I.

103

1. The perpendiculars drawn to the sides of a triangle from their

middle points are concurrent.

Let ABC be a A, and X, Y, Z the

middle points of its sides :

then shall the perp^ drawn to the

sides from X, Y, Z be concurrent.

From Z and Y draw perps to A B, AC;

these perps, since they cannot be parallel,

will meet at point O. Ax. 12.

Join OX.

Because

Tt is required to prove that OX is perp. to BC

Join OA, OB, OC.

Inthe AÂ« OYA, OYC,

YA = YC,

and OY is common to both ;

I also the z OYA = the Z OYC, being rt.L\

.-. OA = OC.

Similarly, from the a^ OZA, OZB,

it may be proved that OA = OB.

Hence OA, OB, OC are all equal.

Again, in the aÂ« OXB, OXC

( BX = CX,

Because <and XO is common to both ;

( also OB = 00:

.-. the Z OXB = the Z OXC;

but these are adjacent z ^ ;

.-. they are rt. l^;

that is, OX is perp. to BC.

Hence the three perps OX, OY, OZ meet in the point O.

Q. E

2. The bisectors of the angles of a triangle are concurrent.

Let ABC be a A . Bisect the Z Â« ABC, A

BOA, by straight lines which must meet

at some point O. Ax. 12.

Join AO.

It is required to prove that AO bisects the

Z BAG.

From O draw OP, OQ, OR perp. to the

sides of the a .

Then in the a** OBP, OBR,

I the Z OBP = the Z OBR,

Because .| and the z OPB = the z ORB, being rt. L^,

( and OB is common ;

.-. OP = OR.

Hy2).

1.4.

Hyp-

Proved.

1.8.

Def. 7.

I. 20.

104

EUCLID'S ELEMENTS.

Similarly from the a" OCP, OCQ,

it may be shewn that OP=OQ,

.-. OP, OQ, OR are all equal.

Again in the a" ORA, OQA,

I the z" ORA, OQA are rt. l',

P land the hypotenuse OA is

liecause. common,

[ also OR -OQ; Proved.

.'. the z RAO = the z QAO. Ex. 12, p. 91.

That is, AO is the bisector of the z BAG.

Hence the bisectors of the three z ** meet at the point O.

Q. E. r>.

3. The bisectors of two exterior angles of a triangle and tlie

bisector of the third angle are concurrent.

Let ABC be a a, of which the sides AB,

AC are produced to any points D and E.

Bisect the Z^ DBC, ECB by straight lines

which must meet at some point O. Ax. 12.

Join AO.

It is required to inove that AO bisects the

angle BAC.

From O draw OP, OQ, OR perp. to the

sides of the a .

Then in the aÂ« OBP, OBR,

the Z OBP = the z OBR, Gomtr.

also the z OPB = the z ORB,

being rt. l ",

and OB is common ;

.-. OP = OR.

Similarly in the a" OCP, OCQ,

it may be shewn that OP = OQ:

.-. OP, OQ, OR are all equal.

Again in the A' ORA, OQA,

( the Z " ORA, OQA are rt. l %

Because \ and the hypotenuse OA is common,

( alsoOR = OQ;

.-. the z RAO = the Z QAO.

Because

Proved.

Ex. 12, p. 91.

That is, AO is the bisector of the z BAC.

.-. the bisectors of the two exterior Z " DBC, ECB,

and of tlie interior z BAC meet at the point O.

Q.B.D.

THEOREMS AND EXAMPLES ON BOOK I.

105

4. The medians of a triangle are concurrent.

Let ABC be a a . Let BY and CZ be two of its

medians, and let them intersect at O.

Join AO,

and produce it to meet BC in X.

It is required to shew that AX is the remaining

median of the a .

Through C draw CK parallel to BY:

produce AX to meet CK at K.

Join BK.

In the A AKC,

because Y is the middle point of AC, and- YD is

parallel to CK,

. â€¢. O is the middle point of AK. Ex. 1, p. 96.

Again in the a ABK,

since Z and O are the middle points of AB, AK,

.-. ZO is parallel to BK, Ex. 2, p. 96.

that is, OC is parallel to BK :

.-. the figure BKCO is a par".

But the diagonals of a parâ„¢ bisect one another, Ex. 5, p. 64.

.-. X is the middle point of BC.

That is, AX is a median of the a .

Hence the three medians meet at the point O. q.e.d.

Corollary. The three medians of a triangle cut one another at a

point of trisection, the greater segment in each being toioards the

angular point.

For in the above figure it has been proved that

AO = OK,

also that OX is half of OK;

.-. OX is half of OA :

that is, OX is one third of AX.

Similarly OY is one third of BY,

and OZ is one third of CZ. q.e.d.

By means of this Corollary it may be shewn that in any triangle

the shorter median bisects the greater side.

[The point of intersection of the three medians of a triangle is

called the centroid. It is shewn in mechanics that a thin triangular

])late will balance in any position about this point : therefore the

centroid of a triangle is also its centre of gravity.]

106

EUCLIDS ELEMENTS.

5. 21ie perpendiculars drawn from the verticeit of a triaiiaJ<' to tJir

opposite sides are concurrent.

Let ABC be a A, and AD, BE, CF the three perp" drawn from

the vertices to the opposite sides :

then shall these perp* be concurrent.

Through A, B, and C draw straight lines MN, NL, LM parallel

to the opposite sides of the a .

Then the figure BAMC is a par'". Bef. 26.

.-. AB=MC. 1.34.

Also the figure BACL is a par".

.-. AB=LC,

.-. LC = CM :

that is, C is the middle point of LM.

So also A and B are the middle points of M N and N L.

Hence AD, BE, CF are the perp* to the sides of the a LMN from

their middle points. Ex. 3, p. 54.

But these perp' meet in a point: Ex. 1, p. 103.

that is, the perp* drawn from the vertices of the a ABC to the

opposite sides meet in a point. q.e.d.

[For another proof see Theorems and Examples on Book iii.]

Definitions.

(i; The intersection of the perpendiculars drawn from the

vertices of a triangle to the oi)posite sides is called its ortho-

centre.

(ii) The triangle formed b}^ joining the feet of the perpen-

diculars is called the pedal triangle.

THEOREMS AND EXAMPLES ON BOOK I.

107

K

VII. ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS.

No general rules can be laid down for the solution of

problems in this section; but in a few typical cases we give

constructions, which the student will find little difficulty in

adapting to other questions of the same class.

1. Construct a right-angled triangle, having given the hypotenuse

and the sum of the remaining sides.

[It is required to construct a rt. ~

angled A , having its hypotenuse equal

to the given straight line K, and the sum

of its remaining sides equal to AB.

From A draw AE making with BA

an z equal to half a rt. l . From

centre B, with radius equal to K, de-

scribe a circle cutting AE in the points

cc.

From C and C draw perp' CD, CD' to AB; and join CB, C'B.

Then either of the a^ CDB, C'D'B will satisfy the given conditions.

Note, If the given hypotenuse K be greater than the perpendicu-

lar drawn from B to AE, there will be two solutions. If the line K be

equal to this perpendicular, there will be one solution ; but if less, the

problem is impossible.]

2. Construct a right-angled triangle, having given the hypotenuse

and the difference of the remaining sides.

3. Construct an isosceles right-angled triangle, having given the

sum of the hypotenuse and one side.

4. Construct a triangle, having given the perimeter and the angles

at the base.

R B

J

[Let AB be the perimeter of the required a , and X and Y the

the base.

From A draw AP, making the / BAP equal to half the z X.

From B draw BP, making the Z ABP equal to half the z Y,

From P draw PQ, making the Z APQ equal to the z BAP.

From P draw PR, making the z BPR equal to the z ABP.

Then shall PQR be the required a .]

Z%at

108 EUCLID'S ELEMENTS.

5. Coustruct a right-angled triangle, having given tlie perimeter

and one acute angle.

6. Construct an isosceles triangle of given altitude, so that its

base may be in a given straight line, and its two equal sides may pass

through two fixed points. [See Ex. 7, p. 49.]

7. Construct an equilateral triangle, having given the length of

the perpendicular drawn from one of the vertices to the opposite side.

*- 8. Construct an isosceles triangle, having given the base, and

the difference of one of the remaining sides and the perpendicular

drawn from the vertex to the base. [See Ex. 1, p. 88.]

9. Construct a triangle, having given the base, one of the angles

at the base, and the sum of the remaining sides.

10. Construct a triangle, having given the base, one of the angles

at the base, and the difference of the remaining sides. ,

Nil. Construct a triangle, liaving given the base, the difference

of the angles at the base, and tJie difference of the remaining sides.

k

[Let AB be the given base, X the difference of the z ' at the base,

and K the difference of the remaining sides.

Draw BE, making the Z ABE equal to half the Z X.

From centre A, with radius equal to K, describe a circle cutting BE

in D and D'. Let D be the point of intersection nearer to B.

Join AD and produce it to C.

Draw BC, making the z DBC equal to the z BDC.

Then shall CAB be the a required. Ex. 7, p. 101.

Note. This problem is possible only when the given difference K

is greater than the perpendicular drawn from A to BE.]

~12. Construct a triangle, having given the base, the difference of

the angles at the base, and the sum of the remaining sides.

"HES. Construct a triangle, having given the perpendicular from the

vertex on the base, and the difference between each side and the

adjacent segment of the base.

THEOREMS AND EXAMPLES ON BOOK I. 109

14. Construct a triangle, having given two sides and the median

which bisects the remaining side, [See Ex. 18, p. 102.]

15. Construct a triangle, having given one side, and the medians

which bisect the two remaining sides.

[See Fig. to Ex. 4, p. 105.

Let BC be the given side. Take two-thirds of each of the given

medians; hence construct the triangle BOC. The rest of the con-

struction follows easily.]

16. Construct a triangle, having given its three medians.

[See Fig. to Ex. 4, p. 105.

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