PROP. 19, THEOR.

If a right line (DC) be drawn bisecting any angle (ADB) of a scelene triangle so as cut the opposite side (AB) the segment (AC) of the cut side next the lesser angle is greater than the segment next the greater angle.

Fig 17.

ANALYSIS.

Since the angle DBA is greater than DAB the side DA is greater than DB: make DE to to DB and join EC; the triangles DEC and DBC are = in every way. But since AC is greater than CB and ... than EC the angle AEC is greater than EAC; this then is evidently the thing to be proved, that the prop. might be proved. Therefore if you make DE = DB and join EC the triangles DEC, DBC are in every way; but the angle AEC is evidently greater than ECD and.. than DCB; but DCB is greater than DAC.. CEA is much greater than CAE for which reason the line AC is greater than EC and... than CB.

=

PROP. 20 PROB.

To find a point within a given triangle (CHF) from which straight lines drawn to the several angles will divide the triangle into three equal parts.

Suppose I to be the point and that the lines IC, IH and IF bisect the triangle.

Through I draw IA, ID parallel to HC, HF; join HA and HD. Since AI is parallel to CH the triangle CIH is to CAH but CIH is of the entire ... CAH is for the same reason FDH is... ADH is ... the three triangles CAH, ADH and DFH are and of ..the bases CA, AD and DF are =.

=

altitudes

Therefore if the base CF be trisected and the lines AI and DI be drawn respectively par. to CH and FH, it is evident that the point in which they meet is the required point; for the equality of the triangles on the same base and between the same parallels can be proved as in analysis... &c. &c.

PROP. 21, THEOR.

If right lines (AE, CE) be drawn from the extremities of one side (AC) of triangle to any point within it; they form with the other two sides angles the sum of which is equal to the difference of the vertical angle (ABC) of the triangle and the angle AEC contained by the drawn lines.

Fig. 19.

Draw BE and produce it to D: the angle AED is to the angles EBA and EAB.. EAB is the difference between AED and ABC, in like manner ECB is the difference between DEC and DBC.. &c.

PROP. 22, PROB.

Given an isosceles triangle (ABC) to construct another on the same base with it (AC) having double its vertical angle. Fig. 19.

ANALYSIS.

Suppose the angle AEC double the angle ABC, join BE and produce it to D. Then because AEC is isosceles the sides AE, EC are, and also the sides A B and BC are = and BE common to the triangles AEB and CEB (by 8th 1 of Elr.) those triangles are equiangular.. the angle ABC is bisected by BD; also the angles ABE, EAB are together to CBE, ECB.. the angle AED (which is to one pair) is to CED (that is to the other pair).. AEC is bisected.. AED is double of ABE.. the angles ABE and BAE are in like manner it can be proved that CBF and ECB

=

are.

Therefore it is evident, that if the angle ABC be bisected by BD and that if the angles BAE and BCE be made each to half ABC that the angle AEC will be double of ABC. For it is to the three angles ABC, BAE and ECB, but they are together double of ABC, .:. &c.

=

NOTE. This does not hold when the vertical angle is right.

PROP. 23. THEOR.

If any angle (BAC) of a tingle, be bisected by (AL) and a perpendicular (AE) be let fall from this angle

on the opposite side, the angle (LAE) contained by bisecting line and perpendicular, is equal to half the difference between the other two angles of the triangle.

Fig. 16.

Make AD to AC, join DC.

=

Then ADC is an isosceles triangle and the angle DAC is bisected by AG.. the angle AGD is a right angle (by Cor. 10. 1. Elr.) and to AEB; but AGD is to AKG and GAK (16. 1. Elr.) and AEB is to ECK, EKC, but EKC is to AKG.. KAG is = to ECK, but ECK is to the difference of the angles ACB, ABC (prop. 18 of dedu.).. &c. &c.

PROP. 24. THEOR.

Right lines (EA,CF) drawn from the extremities of one side (CE) of a triangle, to the middle points of the opposite sides, cut one another in a point of trisection.

Suppose FI to be of FC, it is... of IC: produce it until IG is to IC; join EG, and draw DB through the point of intersection I, then the angles IFD, EFG are= (by 15. 1. Elr.) and the sides about them are respectively

(by constr.) .. the angles FEG,FDI are but they are alternate.. EG is parallel to BD.

Therefore it is evident if DB be drawn, and EG parallel to it, and CF produced to G, that FG is to FI (26. 1. Elr.) but since CB is to BE, and BI par. to EG, CI is to IG (prop. 6. of Anal.) ... IF is į IC and. of CF. In like manner it can be proved that AI is of AE.

PROP. 25. THEOR.

If lines (CF,EA) be drawn from any two angles of a triangle to bisect the opposite sides, the line (DB) drawn from the other angle through their point of intersection (1) bisects the third side.

Fig. 20.

Produce CF until FG is to IF and join GE.

Then it is evident from (prop. 24. of Anal.) that CI is to IG, and that GE is parallel to IB.. (by prop. 6. of Anal.) CB is to CE.

PROP. 26. THEOR.

If a right line (BD) bisect the vertical angle and base of a triangle, it is isosceles.

Fig. 21.

Produce BD until DE is to it join EC.

Then in the triangles BDA,EDC the angles ADB, CDE are and the sides about them are respectively = .. CE is to BA and the angle ABD is to the angle DEC.. DEC is to CBD .. EC is to CB (prop. ... AB is to

6. 1. Elr.) but EC is

to AB (by dem.)

CB.. the triangle is is isosceles.

PROP. 27. PROB.

To subtend a given rectilineal angle (BAC) with a given right line, making an angle equal to a given

Suppose it done, and that BC is to the given line, and the angle BCA to the given one at which BC is to be drawn; draw CD and parallel to BA and join AD. Then it is evident that AD is and parallel to BC, and that the angle BCA is = to CAD.

Therefore, if from the vertex A of the angle to be subtended, AD is drawn to the given line, and making the angle CAD to the angle at which the line is to be drawn; it is evident if DC be drawn par. to AB, and CB par. to DA, that CB is to the given liue, and ACB to the given angle.

PROP. 28. PROB.

Given of a right angled triangle the hypothenuse (BC) and point (D) where perpendicular falls to construct it.

Suppose BGC to be the required triangle. Bisect BC in E, join EG, it is to EB or EC.

Therefore if you raise a perpendicular DG from the given point and bisect BC in E, and subtend the angle BDG with EG drawn from the point of bisection and join BG and CG: BGC is the required triangle. For the angle BGE is to GBC, and also the angle CGE to ECG. .. the angle BGC is = to the two GBC,GCB, and .. CGB is a right angle (cor. 1. 32. 1. Elr.)_

PROP. 29. PROB.

Given the perpendicular of an equilateral triangle to construct it.

Fig. 24.

Draw any infinite right line AB through the extremity of the given perpendicular CD, and at right angles to t; through the other extremity D draw DE and DF making the angles EDC, FDC each = to f the angle of an equi lateral triangle; DEF is the required triangle.

For since CDF is an angle of 30 degrees, and DCF an angle of 90, CFD must be an angle of 60, .. the angle of an equilateral triangle, .. DEF must be an angle of an equilateral triangle.

PROP. 30. THEOR.

The rectangle under the hypothenuse and perpendicular of a right angled triangle, is equal to the rectangle under the sides.

Fig. 25.

For complete the rect. DC under the hypoth. and perpendicular, and also the rect. BF under the sides: then it is evident that the given triangle ABC is of each,.. they are =.

PROP. 31, PROB.

Given (BC) equal to the sum of one side and hypothenuse of an isosceles right angled triangle to construct it,