Lawrence S. (Lawrence Sluter) Benson.

# Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson online

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Font size 2. The difference BED is equivalent (T. 20) to the two angles
ECD, EDC, which (I. 1, cor.) are equal ; therefore ECD is half
of BED. Again : in the triangles AlID, AIIC, because AD,
AC are equal, AIT common, and the contained angles equal,
DII is equal (I. 3) to IIC, and the angles at H are equal, and
are therefore right angles. Then, in the triangles AEF, CEII,
the angles AFE, CUE are equal, and AEC common ; therefore

EXERCISES. 229

(T. 20, cor. 5) the article EAF is equal to ECU, which has been
proved to be equivalent to the half of BED,

3, Since the anole BAG is common to the triangjles ABC,
ADC, the angles ADC, ACD are (I. 20) equal to ABC, ACB ;
and therefore, since ADC, ACD are equal, each of them is
equivalent to half the sum of ABC, ACB ; also either of them,
ADC, is the complement ofDAIi, half the vertical angle, since
AHD is a right angle.

4. Because DH, HC are equal, and HE, DG parallel, GE is
equal (V. 2) to EC; and therefore BG, the difference of BE,
GE, is also the difference of BE, EC.

Sclw. 1. It is easy to prove without proportion, that if AB (in
the figure for the above proposition) be bisected in D, the
straight line DE parallel to BC bisects AC, and that the trian-
gle ADE is a fourth of ABC. For (I. 15, cor. 5) the triangles
BDC, BEC are equal. But (I. 15, cor. 5) BDC is half of ABC ;
and therefore BEC is half of ABC, and is equal to BEA.
Hence the bases AE, EC are equal ; for if they were not, the
triangles ABE, CBE (I. 15, cor. 6) would be unequal. Again :
because AD, DB are equal, the triangle ADE is (I. 15, cor. 5)
half of ABE, and therefore a fourth of ABC.

Conversely, if DE bisect AB, AC, it is parallel to BC. For
(T. 15, cor.) the triangles BDC, BEC are each half of ABC;
and these being therefore equal, DE is parallel (I. 15, cor. 5) to
BC.

Hence, it is plain (I. 14) that the straight lines joining DE,
DF, EF divide the triangle ABC into four equal triangles,
similar to the whole and to one another; and that each of
these lines is equal to half the side to which it is parallel.

Scho. 2. Instead of cutting off AD equal to AC, AC may be
produced through C, and by cutting off, on AC thus produced, a
part terminated at A, and equal to AB, and by making a con-
struction similar to that of the foregoing proposition, it will be
easy to establish the same properties as those above demon-
strated, or ones exactly analogous.

Prop. IX. â€” Prob. â€” Giveii the base of a triangle, the differ-
ence of the S'des, and the difference of the angles at the base;
to construct it.

230 EXERCISES.

Make BC equal to the given base, and CBD equal to half
the difference of the angles at the base ; from C as center, at a
distance equal to the difference of the sides, describe an arc
cutting BD in D ; join CD and produce it ; make the angle
DBA equal to BDA ; ABC is the required triangle.

For (I. 1, cor.) AD is equal to AB, and the difference of AC,
AD, or of AC, AB, is CD ; and (Ex. 8)
since AD is equal to AB, CBD is equal
to half the difference of the angles at the
base. The triangle ABC, therefore, is
the one required, as it has its base equal
to the given base, the difference of its
sides equal to the given difference, and
the difference of the angles at the base equal to the given dif-
ference of those angles.

Method of C\miputation. In the triangle BCD there are
given BC, CD, and the angle CBD ; whence the angle C can
be computed ; and the sum of this and twice CBD is the angle
ABC. Then, in the whole triangle ABC, the angles and the
Bide BC are giveti ; whence the other sides may be computed ;
or, one of them being computed, the other will be found by
means of the given difference CD.

Prop. X.â€” Prob. â€” Given the segments into which the base
of a triangle is divided by the line bisecting the vertiecd anglQ
and the difference of the sides ; to construct the triangle.

Construct the triangle CED, having the sides CE, ED equal
to the given segments, and CD equal to the given dillerence of
the sides; produce CE, and make EB equal to ED; bisect the
angle BED by EA, meeting CD produced in A, and join AB ;
ABC is the required, triangle.

A For, in the triangles AEB, AED, BE

is equal to ED, EA common, and the
angle BEA equal to DEA ; therefore (I,
3) BA is equal to DA, and the angle
EAB to EAD. Hence, ABC is the re-
quired trianale; for CD, the difference
of its sides, is equal to the given differ-
ence, and BE, EC, the segments into which the base is

FXERCISE8.

231

divided by the line bisecting the vertical angle, are equal to
the given segments.

Meth d of Computation. The sides of the triangle CDE
are given, and therefore its angles may be computed; one of
which and the sni)plement of the other are tlie angles C and
B. Then, in the triangle ABC, the angles and BC are given,
to compute the remaining sides.

OTHERWISE,

Since (V. 4) CE : EB : : CA : AB, we have, by division,
CEâ€” EB : EB : : CAâ€” AIJ : AB ; which, therefore, becomea
known, since the first three terms of the analogy are given;
and thence AC will be found by adding to AB the given dit-
ference of the sides.

Prop. XT. â€” I V.ob. â€” Given the base of a triangle^ the vertical
angle, and the difference of the sides ; to construct the tri-
angle.

Let MNO be the given vertical angle; produce ON" to P,
and bisect the angle MNP by NQ. Then, make BD equal to
the difference of the sides, and the
angle ADC equal to QNP; from B
as center, with the given base as ra-
dius, describe an arc cutting DC in
C ; and make the angle DCA equal
to ADC ; ABC is the required tri-
angle. For (const.) the angles ACD,
ADC are equal to MNP, and there-
fore (I. 20 and 9) the angle A is
equal to MN). But (I. 1, cor.) AD
is equal to AC, and therefore BD is
the difference of the sides AB, AC ;
and the base BC is equal to the
given base ; wherefore ABC is the triangle required.

Method of Computation. In the triangle CBD, the sides
BC, BD, and the angle BDC, the supplement of ADC or MNQ
are given ; whence the other angles can be computed. The
rest of the operation will proceed as in the ninth proposition of
these Exercises.

Q

N

232 EXERCISES.

Prop. XII. â€” Prob. â€” Given one of the angles at the base of
a triangle, the difference of the sides, and the difference of the
segments into which the base is divided by the line bisecting the
vertical angle; to construct the triangle.

Construct the triangle DBG, liaving the angle B equal to
the given angle, BD equal to the difference of the sides, and

BG equal to the difference of the segments ;
draw DC perpendicular to DG, and meet-
^'^ '^ ing BG produced in C; produce BD, and

make the angle DCA equal to CDA ;
B G E c ABC is the triangle required. For it has
B equal to the given angle, and the differ-
ence of its sides BD equal to the given difference; and if AHE
be drawn bisecting the angle BAC, it bisects (I. 3) CD, and is
perpendicular to it ; it is therefore parallel to DG, one side of
the triangle CDG; and, bisecting CD in II, it also (V, 2)
bisects GC in E. Hence BG, the difference of BE, GE is also
the difference of BE, EC, the segments into which the base is
divided by the line bisecting the vertical angle.

Method of Computation. In the triangle DBG, BD, BG,
and the angle B are given ; whence (Trig. 3) we find half the
difference of the angles BGD, BDG, which is equal to half the
angle C. Then (by the same proposition) we have in the tri-
angle ABC, tan I (C-B) : tan ^ (C+B) : : c-b or BD : c+
b ; whence the sides c and b become known, and thence BC
by the first case.

For (I. 20) BEA=r| A+C, and consequently BEAâ€” | A = C.
But (I. 16) BGD = BEA, and BDG^BAE^^A; and there-
fore BGDâ€” BDG =C.

Prop. XIII. â€” Prob. â€” Given the base of a triangle, the verti-
cal angle, and the sutn of the sides ; to construct it.

Make BD equal to the sum of the sides, and the angle D
equal to half the vertical angle ; from B as center, with the
base as radius, describe an arc meeting DC in C ; and make
the angle DCA equal to D ; ABC is the required triangle.

For (I. 1, cor.) AD is equal to AC, and therefore BA, AC are
equal to BD, the given sum. Also (I. 20) the exterior angle
BAC is equal to the two D and ACD, or to the double of D,

EXERCISES.

233

because D and ACD are equal ; therefore, since D is half the

given vertical angle, BAG is equal

to that ansjle. The triancjle ABC,

therefore, has its base equal to the

given base, its vertical angle equal

to the given one, and the sum of its

sides equal to the given sum ; it is

therefore the triangle required.

Scho. Should the circle neither cut nor touch DC, the prob-
lem would be impossible with the proposed data. If the circle
meet DC in two points, there will be two triangles, each of
which will answer the conditions of the problem. These tii-
angles, however, will differ only in position, as they will be on
equal bases, and will have their remaining sides equal, each to
each. This problem might also be solved by describing (III.
19) on the given base 15C a segment of a circle containing an
angle equal to half the vertical angle ; by inscribing a chord
BD equal to the sura of the sides; by joining DC; and then
proceeding as before. The construction given above is prefer-
able. /

Method of Computation. In the triangle BDC, the angle
D, and the sides BC, BD are given ; whence the remaining
angles can be compute 1 ; and then, in the triangle ABC, the
angles and the side BC are given, to compute the other sides.

Prop. XIV. â€” Prob. â€” Given the vertical angle of a triangle^
and the segments into which the line iisectiiig it divides the
base ; to construct it.

In the straight line BC, take BH and CII equal to the seg-
ments of the base ; on BC describe (III.
19) the segment BAC containing an angle
equal to the vertical angle, and complete
the circle ; bisect the arc BEC in E ; draw
EHA, and join BA, CA ; ABC is the re-
quired triangle. For (III. 16, cor. 1) the
angles BAII, CAH are equal, because the
arcs BE, EC are equal ; and therefore the
triangle ABC manifestly answers the conditions of the ques-
tion.

234 EXEECISF.8.

Scho. The construction might also be effected by describing
on BH and CH segments each containing an angle equal to half
the vertical angle, and joining their point of intersection A
â– with B and C. Another solution may be obtained by the
principle (V. 4) that BA â€˘ AC : : BII : HC. For if a trian-
gle be constructed having its vertical angle equal to the given
one, and the sides containing it equal to the given segments,
or having the same ratio, that triangle will be similar to the
required one ; and therefore on CB construct a triangle equi-
angular to the one so obtained.

Method of Computation. Join BE, and draw ED perpen-
dicular to BC. Then BD or DC is half the sum of the sear-
nients BII, HC, and DH half their difference ; and BD is to
DM, or twice BD to twice DH, as tan DEB to tan DEH.
Now it is easy to show that BED is half the sum of the angles
ABC, ACB, and HED half their difference ; and therefore these
angles become known; and BC being given, the triangle ABC
is then resolved by the method for the Hrst case.

Cor. Hence, we have the method of solving the problem in
wliich the base, the vertical angle, and the ratio of the sides of
a triangle are given^ to construct it. For (V. 4) the sides be-
ing proportional to the segments BH, HC, it is only necessary
to divide the given base into segments proportional to the
sides and then to proceed as above. *

Prop. XV. â€” Pbob. â€” Given the base^ the perpendicular^ and
the vertical angle of a triangle / to construct it.

Make BC equal to the given base, and (HI. 19) on it de-
scribe a segment capable of containing
an angle equal to the vertical angle ;
draw AK parallel to BC, at a distance
from it equal to the given perpendicular
and meeting the arc in A ; join AB,
AC; A13C is evidently the tiiangle re-
quired.

Method of Compiitat'on. Draw the

perpendicular AD, and parallel to it

draw LGH, through the center G; join BG, AG, AH. Now,

iiuce AH evidently bisects the anglu BAC, the angle HAD or

EXERCISES.

235

n is (Ex. 8) equal to half the difference of the ansrles ABC,
ACB, and therefore (III. 1 0) AGK is the wliole difference of
those angles. Then, in the riglit-angled triangle BP'G, the
angles and BF being known, FG can be computed ; from
which and from AD or KF, KG becomes known. Now, to the
radius BG or AG, FG is the cosine of BGF, or BAG, and KG
the cosine of AKG; and therefore FG : KG : : cos liAC :
cos AGK. Hence the angles ABC, ACB become known, and
thence the remaining sides.

iSc/io. Should the parallel AK not meet the circle, the solu-
tion would be impossible, as no triangle could be constructed
having its base, perpendicular, and vevlic:il angle of the given
magnitudes. If the parallel cut the cinrle, there will be two
triangles, either of which will answer the conditi(m of the ques-
tion. They will differ, however, only in position, as their sides
will be equal, each to each. If the parallel touch the circle,
there will be only one triangle ; and it will be isosceles.

Pkop. XVI. â€” Prob. â€” Given the base of a triangle, the ver-
tical angle, and the radius of the inscribed circle ; to construct
the triangle.

Let GHK be the given angle ; produce GH to L, and bisect
LHK by HM; on the given base BC desciibe (III. 19) the seg-
ment BDC, containing an angle equal to GHM ; draw a straight
line parallel to BC, at a distance equal to the given ra<lius, and
meeting the arc of the segment in D; join I)B, DC; and make
the angles DBA, DCA equal to DBC, DCB, each to each ;
ABC is the required triangle. L ii g

Produce BD to E, and drawDF perpen-
dicular to BC. Then, since (const.) the
angle BDC is equal to GHM, the two jj
DBC, DCB are equal (I. 20 and 9) to
LHM, and therefore (const.) ABD, ACD
are equal to KHM. But (1. 20) BDC is
equal to BEC, ECD, or to BAC, ABD,
ACD, because (I. 20) BEC is equal to
BAC, ABD. Therefore BAC, ABD, ACD
are equal to GHM; from the former take
ABD, ACD, and from the latter KHM, which is equal to

236

KXERCISE8.

them, and the remainders BAG, GHK are equal. It is plain,
also (I. 14), tliat perpendiculars drawn IVoni D to AB and AC
would be each equal to DF ; and therefore a circle described
from D as center, with DF as radius, would be inscribed in the
trian<ijle ABC ; and BC being the given base, and A being
equal to the given vertical angle, ABC is the required tri-
angle.

The method of computation is easily derived from that of the
preceding proposition.

Scho. Should the parallel to BC not meet the arc of the seg-
ment, the solution would be impossible, as there would be no
triangle which could have its base, its vertical angle, and its
inscribed circle of the given magnitudes. If the parallel be a
tangent to the arc, the radius of the inscribed circle would be
a maximum. Hence, to solve the problem in which the base
and the vertical angle are given, to construct the triangle, so
that the inscribed circle may be a maximum, describe the seg-
ment as belbre, and to find D bisect the arc of the segment.
The rest of the construction is the same as before ; and the tri-
angle will evidently be isosceles.

Prop. XVII. â€” Prob. â€” Given the three lines drawn from the
vertex of a triangle, one of them 'perpend cnlorto the base, one
bisecting the base, and one bisecting the vertical angle y to con-
struct the triitngle.

Take any straight line BC and draw DA perpendicular to it,
and equal to the given perpendicular; fi-om A as center, with
radii equal to the lines bisecting the vertical angle and the

base, describe arcs cutting BC in E and
F, and draw AEH and AF ; through F
draw GFII perpendicular to BC, and
draw AG making the angle IIAG equal
to II, and cutting HG in G; from G as
center, with GA as radius, describe a
circle cutting BC in B and C; join AB,
AC ; ABC is the triangle required.
For (HI. 2) since GFII is perpendicu-
lar to BC, BC is bisected in F ; and (III. 17) the arcs BII, HO
are equal. Therefore (III. 16, cor. 1) the angles BAII, CAH

EXERCISES. 237

are equal. Hence, in the triangle ABC, the perpendicular AD,
the line AE bisecting the veitical angle, and the line AF
bisecting the base, are equal to the given lines. Therefore
ABC is the triangle required.

Scho. It" the three given lines be equal, the problem is inde-
terminate ; as any isosceles triangle, having its altitude equal
to one of the given lines, will answer the conditions.

3fethod of Computation. Through A draw a parallel to BC,
meeting HG ))rodnced in K, Then, in the right-angled triangle
ADE, AE, AD being given, DAE, or H, may be computed ;
the double of which is AGK ; and AK or FD being given, AG,
GK can be found, and thence GF. Hence, if GB were drawn,
it and GF being known, BF, and the angle BGF, or BAC, can
be computed. The rest is easy; DAE, half the difference of B
and C, being known.

Analysis. Let ABC be the required triangle, AD the per-
pendicular, and AE, AF the lines bisecting the vertical angle
and the base. About ABC describe (HI, 25, cor. 2) a circle,
and join its center G, with A and F, and produce GF to meet
the circumference in H. Then (HI. 2) GFH is perpendicular
to BC, and (IH. \1) the arcs BH, HC are equal. But (HI. 16)
the equal angles BAE, CAE at the circumference stand on
equal arcs; and therefore AE being produced, will also pass
through H; and the point H, and the angle GHA and its equal
HAG are given. Hence also the center G and the circle are
given, and the method of solution is plain.

Prop. XVIH. â€” Prob. â€” Given the base of a triangle^ the ver-
tical angle, and the straight line bisecting that angle y to con-
struct the triangle.

On the given base BC describe (HI. 19) the segment BAC
capable of containing an angle equal to the given vertical an-
gle, and complete the circle ; bisect the arc BEC in E, and join
EC; perpendicular to this, draw CF equal to half the line
bisecting the vertical angle, and from F as center, with FC as
radius, describe the circle CGII, cutting the straight line pass-
in through E and F in G and H ; make ED equal to EG, and
draw EDA; lastly, join AB, AC, and ABC is the required
triangle.

238 EXEKCI8E8.

For the triangles CEA, CED are equiangular, the angle CEA
being tomnion, and BCE, CAE being each equal to BAE.
Therefuie AE : EC : : EC : ED, and (V. 9, cor.) AE.ED=

EC=. But (ITI. 16, cor. 3, and 21) HE.EG or HE.ED=EC
and therefore AE.ED^HE.ED; whence AE=:HE, and (I. ax.
3) AD=GH = '2CF. AD is therefore equal to the given bisect-
ing line, and it bisects the angle BAC. Hence ABC is the re-
quired triangle.

Method of Computation. Draw EL perpendicular to BC,
and join CH. Then BCE is equal to BAE, half the vertical
angle A ; and therefore, to the radius EC ; CL is the cosine of
â– Â§ A, and CF is the tangent of CEF to the same radius ; where-
fore, to any radius, CL : CF, or BC : AD : : cos ^ A : tan
CEF or cot EFC ; and hence the angle H, being half of EEC,
is known. Also ECU is the complement of II, because ECF is
a right angle, and FCH equal to H. But (Tgig. 2) EC : EH or
EA : : sin II : sin ECU, or cosH ; or (Trig. defs. cor. 6) EC :
EA : : It : cot H. Also in the triangle ACE, EC : EA : :
Bin -^ A : sin ACE ; whence (IV. 7) R : cot H : : sin ^ A : sin
ACE ; whence ACE may be found ; and if from it, and from
ABE, its supplement (BE being supposed to be joined) BCE
be taken, the remainders are the angles at the base.

Analysis. Let ABC be the required triangle, and let AD,
the line bisecting the vertical angle, be produced to meet the
circumference of the circumscribed circle in E; join also EC.
Then (III. 19) the circumscribed circle is given, since the base
and vertical angle are given ; and the arc BEC is given, as are
also its half EC, and the chord EC. Now the triangles AEO,

EXERCISES. 239

DEC are equiangular ; for the angle CEA is common, and (IH.
18, cor. 1) BCE is equal to BAE or EAC. Hence AE : EC : :
EC : ED, and theiefore AE.ED=ECl Hence (IH. 21) it is
evident that if EC be made a tangent to a circle, and if
through the extremity of the tangent a line be drawn cutting
the circle, so that the part within the circle may be equal to
AD, DE will be equal to the external part ; whence the con-
struction is manifest.

Prop. XIX. â€” Proh. â€” Given the straight liius drawn from
the three angles of a trian;^ le to the points of bisection of the
opposite sides / to construct the triangle.

Trisect the three given lines, and describe the triangle ABC
having its three sides respectively equal to two thirds of the
three given lines ; complete the parallelogram ABEC, and
draw the diagonal AE ; produce also
CB, making BE equal to BC ; and
join FA, FE; AFE is the required
triangle.

Produce AB, EB to G, H. Then
(H. 14, cor.) AE, BC bisect each other
in D, and therefore FD is equal to
one of the given lines, for BD is one

third of it, and FB two thirds. Again : because FB, BC are
equal, and HB pai-allel to AC, FA is bisected in H, and HB is
half of AC or BE. Hence, HE is equal to another of the given
lines, and it bisects FA. In the same manner it would be
proved, that AG is equal to the remaining line, and that it
bisects FE. Hence FAE is the triangle required.

Method of Computation. BD, which is a third of one of
the given lines, bitects AE, a side of the triangle ABE, in
â€˘which the sides AB, BE are respectively two thirds of the two
remaining lines. Then (I. 24, cor. 6, and II. 12, cor.) 2AD^=
AB'+BE*â€” 2BD'; whence AD, and consequently AE may be
found ; and in the same manner the other sides may be com-
puted.

Prop. XX.â€” Prob, â€” Given the three perpendiculars of a tri-
angle y to construct it.

240

EXERCI8E8.

A-
B-
C-
D-

E

Let A, B, C be three given straight lines; it is required to
describe a triangle having its three perpendiculars respectively
eqnJil to A, B, C.

Take any straight line D, and describe a triangle EFG. hav-
ing the sides FG, P'E, EG third i)ro-
portionals to A and 1), B and D, C
and D; and draw the perpendiculars
EH, GL, FK.

Then the rectangles rG.A,EF.Bare
equal, each being equal to the square
ofD; and therefore EF : FG : : A :
B. But in the similar triangles EHF,
GLF, EF : FG : : EH : GL; where-
fore EH : GL : : A : B ; and in the
same manner it would be proved, that
EH : FK : : A : C. Hence (IV. 7)
if EH be equal to A, GL is equal to B, and FK to C ; and
EFG is the triangle requii-ed.

But if EH be not equal to A, make EM equal to it, and draw
NMO parallel to FG, and meeting EF, EG, produced, if neces-
sary, in N and O ; ENO is the required triangle. Draw OP
perpendicular to EN. Then EM : EH : : EO : EG, and OP
GL : : EO : EG ; whence (IV. 7 and alternately) EM : OP :
EH : GL ; or, by the foregoing part, EM : OP : : A : B
wherefore (IV. 7) since EM is equal to A, OP is equal to B
and it would be proved in a similar manner, that the perpen-
dicular from N to EO is equal to C.

Method of Computation. By dividing any assumed num-
ber successively by A, B, C, we find the sides of the triangle
EFG, and thence its angles, or those of ENO; whence, since
the perpendicular EM is given, the sides are easily found.

Or, when the sides of EFG are found, its perpendicular EH
may be comj)Uted in the manner pointed out in the note to tlie
twelfth proposition of the second book. Then EH : A : : FG:
NO : : EF : EN : : EG : EO.

Prop. XXI. â€” Prob, â€” Given the sum of the legs of a right-
angled triangle, and the sum of the hypothenuse, and the per-
pendicular to it from the right angle; to construct the triangle.

EXERCISES. 241

Let tho sum of the legs of a right-angled triangle be equal
to the straight line A, and the sum of the hypothenuse and
perpendicular equal to BC ; it is required to construct the tri-
angle.

Find (I. 24, cor. 4) a straight line the square of which is equal
to the excess of the square of BC

above that of A, and cut off BD ~

equal to that line ; on DC as diam-
eter describe a semicircle, and draw
EF parallel to DC ate a distance
equal to BD ; join either point of
intersection, E, with D and C ;
DEC is the required triangle.

Draw the perpendicular EG, which (const.) is equal to BD.
Then (II. 4) BC^ or AHEG^ = (DC-fEG)^=DC'^ + EG=+
2DC.EG; whence A^=DC^ + 2DC.EG. Also (DE + EC)^=
DE^ + EC^+2DE.EC=DC^-|-2DC.EG, because (III. 11, and I.
24, cor. 1) DC'=DE^+EC^ and DC.GE^DE.EC, each being
equal to twice the area of the triangle DEC. Hence (DE4-
EC)- = A-; wherefore (I. 23, cor. 3)bE + EC=A; and DEC