Suppose BAD to be the required triangle. Join CA. Then because the triangle is isosceles and the angle BAРa right angle, each of the angles DBA, ADB is a right angle: and since DC is to DA the angles DCA and Dö are = and each of them of a right angle, since they are together to ADB.

=

Therefore it is evident that if from either extremity B, of the given sum, BA be drawn, making the angle CBA to a right angle; and if from the other extremity C, CA be drawn, making the angle BCA of a right angle, and if through A, AD be drawn making CAD = DCA, BAD is the triangle required.

PROP. 32, THEOR.

If the three sides of any triangle (AGC) be bisected, the perpendiculars drawn to the sides from the points bisection, meet in the same point. And the point of concourse is equally distant from the three angles.

Fig. 27.

Draw EI,FI from the points of bisection E,F at right angles to GC and CA: the line drawn from I to B will be at right angles to AG.

=

=

For, join IA,IC and IG; because GE is to EC and IE common and at right angles to GC; GI is to IC: for the same reason IC is to IA ... IG is = to IA. Then because IG is to IA the angles IGB,LAB are = and AB is to GB (by hypoth.) .. the angles GBI, ABI are and are .. right.

=

PROP. 33, PROB.

To determine a point in the side of any triangle (ABC) from which lines drawn parallel to the other sides of the triangle will form an equilateral parallelogram.

Suppose it done, and that IE is an equilateral parallelogram; join. AD. Because IA is to ID the angle

IAD is to IDA; for the same reason the angle EAD to EDA, but the angle EDA is ➡ to DÃI .. the angle EAD is to IAD.

is

Therefore, it is evident that if the angle BAD is bisected, and the line bisecting it be produced to the op posite side, the point D, where it meets it, is the required point. For draw DE and DI par. to BA and AC; the parallelogram IE is equilateral, Because the angle IDA is to DAE and IDA is also to ADE.. IAD is → to IDA, .'. IA is to ID .. &c.

PROP. 34. PROB.

Given the sum of the three sides (AB,) required to con struct a triangle, equiangular to a given one,

Let the given sum be AB; and suppose the triangle DEF to be the required one, and that DE is DA and FE to FB, draw AG par. to DE and BG par. to FE the triangle AGB is equiangular to DEF; join AE and EB, the angle EAG is to AED and AED is to EAD .. the angles DAE and GAE are for the same reason the angles EBG and EBF are

.

Therefore, it is evident, that if you construct on the given sum AB, a triangle AGB equiangular to the giveu one, bisect the angles GAB,GBA by the right lines EA and EB, and draw through the point E in which they meet lines ED and EF par. to GA and GB, the triangle DEF is the required one.

It is evidently equiangular to the given one: and because ED is par. to GA the angles GAE,AED are → but GAE is to EAD (by construction).. the angles DAE,DEA are.. DE is to DA; for the same rea son FB is to FE.. &c. &c.

PROP. 35, THEOR.

If any point (G) be taken in the side of an equilateral triangle (DEF) and from it lines (GI,GK) be drawn parallel to the other sides of the triangle, their sum is equal to a side of the triangle.

F

Fig. 24.

Because GI is par. to DF the angle GIE is angle DFE .. GEI is equilateral.. GI is

to the

to GE, for the same reason GK is to DG .. GK and GI are together to DE.

PROP. 36, THEOR.

If a side (CA) about the bisected angle (CAB) of any triangle be produced until the produced part (AK) is equal to the conterminous side (AB) the right line (KB) joining their extremities will be parallel to the bisector (AD).

Fig. 28.

Because AB is to AK the angle ABK is to the angle AKB ... ABK is half of their sum; and CAB is to their sum, .'. half CAB, viz. DAB is to ABK, but they are alternate, .. KB is parallel to AD.

PROP. 37. PROB.

To draw a right line parallel to the side (AC) of any triangle (ABC) so that the alternate segments (AD,BE) shall be equal.

Suppose it done, and that DE is drawn par. to AC,. making the alternate segments AD and BE=; draw EK par. to BA and join KB

Because EK is par. to DA and ED to KA, EK is = to DA.. to EB.. the angles EBK, EKB are but EKB isto KBA ... KBA is to EBK.

Therefore if the angle ABC be bisected by BK, and KE drawn parallel to AB, the right line drawn from E parallel to AC makes the alternate segments, i. e. BE and DA,. For EK is to DA, and BE is to EK (because the angle BKE as being to ABK is to KBE).. BE is to DA.

1

PROP. 38, PROB.

Given one side (AB) an angle at it, and the difference between the other two sides to construct the triangle.

Suppose ADB to be the required triangle; when the given angle DBA is subtended by the lesser side AD. Cut from DB (the greater) CB to the given difference, join CA. Then AD is to CD .. the angle DAC is to DCA (5. 1. Eir.)

Therefore it is evident, if you draw CB to the given difference, and making with AB the angle ABC = to the given angle, join CA, produce BC, and make the angle CAD to the angle ACD, that ADB is the required triangle.

Fig. 32.

But when the given angle ADB is subtended by the greater side AB, produce BD until CD is difference, join AC. Then since BA is angle BAC is to BCA.

to the given to BC, the

Therefore if you make the angle ADB the given angle, make DC the given difference, join CA, and make the angle CAB BCA, it is evident that ADB is the required triangle.

PROP. 39, PROB.

Given the base (AB) to find the locus of the vertex of an infinite number of triangles equal in area.

Suppose any number of triangles ACB, ADB, AEB, = in area, to be on the given base, join C,D the vertices of any two of them; since the triangles are and on the same base they must be between the same parallels .. CD is par. to AB, and for the same reason DE is par. to AB. it is evident that the locus lies in a line par. to the base.

PROP. 40, THEOR.

The right lines (DE, EF, FG, GD,) joining the middle points of the sides of any quadrilateral figure, form a parallelogram, which is half of the entire figure (ABCI.)

Fig. 34.

Draw IB and AC. Then in the triangle AIC, since the sides AI,IC are bisected, GD connecting the points of section is par. to AC, (prop. 7 of anal.) for the same reason FE is parallel to AC, .. FE is par, to GD; in like manner it can be proved that DE is par. to GF.. FEDG is a parallelogram,

And because AI,IC, are bisected in G and D, the triangle GID is the fourth part of AIC; also FBE is the fourth part of ABC., the triangles GID,FBE are together the fourth part of the entire figure; in like manner it can be proved that the triangles DCE,GAF are together the fourth of the entire figure, .. if all those triangles be taken away, the remaining figure GDEF is half of the entire figure.

PROP. 41, THEOR.

If in any triangle (ABC) a line (BO) be let fall from the vertex perpendicular to the base, and if from the point in which the perpendicular meets the base, lines (OD,OE) be drawn to the points of bisection of the other sides, the triangle (DOE) formed by these lines and the line (ED) joining the points of bisection, is similar to the whole triangle,

Fig. 85.

Because the side BC is biseeted by DI, BO is also bisected by it, and the angles BID,DIO are right, .. BD is to DO, but BD is to DC .. DO is = to DO.. the angles DCO,DOC are, but the angle EDO isto DOC,. to DCO; for the same reason the angle DEO is to CAB,.. the remaining angle DOE is = to the angle ABC,., the triangles BAC,OED ale similar.