PROP. 92. PROB.

Given one angle of any triangle, one side, and the length of the line drawn from the point of concourse of the perpendiculars (from the middle points of the sides) to one of the angles, to construct it.

Suppose that AGC is the required triangle. Let ACG be the given angle, GC the given side, and IC the given line. Draw IF, IE and IB at right angles to the middle points of the sides; join IA and IG. IA and IG are evidently each to IC (prop. 32 of deduc.)

If.. you bisect the given side GC in E, draw EI at right angles to it, draw CI from the given angle = to the given line, and draw (from the point I, where CI meets the perpendicular) IA and IG, each to CI, to meet the legs of the given angle, and join AG: AGC is evidently the required triangle.

PROP. 93. PROB.

Given of a right angled triangle, the sum of the hypotenuse and one side, and the contained angle to construct it.

Suppose DCA is the required triangle. Let BA be = to the given sum, and ACD to the given angle; join DB; the angle DBC is half the given angle, because CB is = to CD; and the angle DAC is the complement of the given angle, it with DCA being to a right angle.

Therefore, if at one extremity A of the given line, you make the angle DAB to the complement of the given angle, and at the other extremity B make the angle ABD = to half the given angle, and make the angle BDC = to DBC; DCA is evidently the required triangle.

For, the angle DCA is to CBD + CDB, and .. to the given angle; DC + CA are to AB, because the angles BDC and CBD are =, and the angle CDA is a right angle, because the angles CAD and ACD are together to a right angle,

PROP. 94. THEOR.

The right line (AD) drawn from the vertex of any triangle (ABC) to bisect the base, is less than half the sum of the sides.

Fig. 28.

Produce CA till AK is to it; join KB; then in the triangle CBK, CB is bisected in D, and CK in A; .. DA is parallel to BK, and to half of BK (prop. 7. of deduc.); but KB is less than the sum of BA and AK; .. than the sum of BA and AC; .. half of BK or AD is less than half of CA and AB.

PROP. 95. PROB.

Given one side, and the perpendicular on the hypotenuse of a right angled triangle, to construct it.

Fig. 25.

=

Draw any right line AC; from any point O in it, draw a line OB at right angles to it, and to the given perpendicular; from its extremity B draw the right line BA, subtending the angle BOA and to the given side, and draw BC, making the angle OBC = to the angle BAC; ... BCA is the required triangle.

=

For, since BOA is a right angle, ABO + BAO are = to a right angle ; ... OBC + OBA are to a right angle; .. the triangle is right angled, and it has the given side and given altitude.

PROP. 96. THEOR.

If the bisection (ICD) of the vertical angle of an isosceles triangle (ACD) be bisected, the intercept (IF) of the base, between the bisectors, is less than the intercept (FD) between the side and second bisector.

Fig. 37.

Draw FG at right angles to CD.

Then, in the triangles CIF and CGF, the angles ICF and GCF are =, also the angles CIF and CGF, and the side CF common to the two triangles; .. FI is to FG (prop. 24. of Elr.); but FG is less than FD, since the angle FDG is less than FGD; ... FI is less than FD.

PROP. 97. THEOR.

If in any parallelogram (AF) right lines be drawn from the opposite angles (FCA and AEF), to the points of bisection (D and G) of their opposite sides (AE and CF), they trisect the diagonal (AF).

Fig. 47.

Since DE and CG are the halves of opposite sides of a parallelogram, they are ; .. CD is parallel to GE. And in the triangle AKE, since AE is bisected in D, and ID parallel to KE, IA is to IK (prop. 6. of dedu.); for the same reason KF is = to KI; ... the three AI, IK and KF are to one another; .. the diagonal is trisected.

PROP. 98. THEOR.

If the right lines (DF and DG) drawn from the middle point of the base of a triangle (BAC), at right angles to the sides, be equal, the triangle is isosceles.

Fig. 19.

Because, in the triangles FAD and GCD, the sides FD and DA of the one are respectively to GD and DC of the other; and the angles DFA and DGC righ angles, it follows (by the 47th 1. of Elr.), that FA and GC are = ; .. by the 8th 1. of Elr. the angles GCD and FAD are, and .. the triangle ABC is isosceles.

PROP. 99. THEOR.

If, from the extremity of the lesser side (AB) of any triangle (ABD), a line. (BC) be drawn, cutting from the greater a part (AC) equal to the lesser, and if the remaining side (BD) be bisected (in E) and a right line (EF) be drawn through the point of bisection, parallel to the cutting line, the distance between the point (F), where it cuts the greater side, and the vertex of the triangle, is equal to half the sum of the greater and lesser sides, and the remainder is equal to half their difference.

Fig. 53.

Because the side BD is bisected in E, and EF drawn

=

parallel to BC, CF is to FD (prop. 6. of deduc.); .. BA and FD together are = to AC and CF; .. AF is half the sum of BA and AD. And since BA is to AC, CD is the difference between BA and AD; .. FD is half the difference.

PROP. 100. PROB.

Given of any triangle, the base, difference of the sides, and vertical angle, to construct it.

Fig. 53.

ANALYSIS.

Suppose that ABD is the required triangle. Let BD be given as its base, CD the given difference, and BAD the given angle; join BC. Because CD is the difference between AB and AD, .. AB is to AC; .. the angles ABC and ACB are, and .. ACB is half the supplement of the vertical angle.

If.. you draw any line, BC; make, at one extremity C of it, an angle BCA = to half the supplement of the vertical angle; produce the right line AC till CD is - to the given difference; from its extremity D inflect the right line DB to the given base, and draw BA, making the angle ABC to ACB; it is evident that ABD is the required triangle.

=

PROP. 101. PROB.

In a right angled isosceles triangle, given the difference between the hypotenuse and side, to construct it.

Suppose that CAD is the required triangle. Let AB be to the given difference; join BC; then DB is to DC, and .. the angle DBC to DCB, and since the triangle CAD is isosceles, the angle CDA is half a right angle; .. the angle CBD is half the supplement of half a right angle.

Then at one extremity (A) of the given difference, make the angle BAC to half a right angle; at the other extremity B, make the external angle DBC = to half the supplement of half a right angle; from the point C, in which the lines making those angles meet, draw CD, making the angle BCD to CBD; the triangle CAD is evidently the required one.

For, since the angles DCB and DBC are =, and each of them half the supplement of half a right angle, CDB is half a right angle, and is.. to DAC; .. ACD must be a right angle; ;. &c.

PROP. 102. THEOR.

The perpendiculars (DF and DG), let fall from the middle point of the base of an isosceles triangle on the sides, are equal.

Fig. 19.

For, in the triangles DFA and DGC, the side DA is = to DC, and the angles DAF and AFD of the one respectively to DCG and CGD of the other; .. the triangles are in every way, and .. DF is to DG.

PROP. 103. PROB.

To draw a right line parallel to the base (AB) of any triangle (GAB), so that the sum of the intercepts between the parallels, shall be equal to the drawn line.

Suppose that HI is the line required to be drawn. Make IE to IB; HE is to HA; join EA and EB. Then, because EI and IB are, the angles IEB and IBE are; and because IH is parallel to BA, the angles IEB and EBA are ; .. the angles IBE and EBA are = ; .'. GBA is bisected; for the same reason GAB is bisected.

If.. the angles at the base be bisected, and the right lines bisecting them be produced to meet, it is evident, that the right line IH, drawn through their point of concourse E, is to the sum of the intercepts HA and IB.

PROP. 104. THEOR.

If right lines (CE, AG, DH and BF) be drawn from the angles of any parallelogram (AD), at right angles to any line (EF), the sum of those (CE and BF), from one pair of opposite angles, is equal to the sum of those (AG and DH) from the other pair of opposite angles. Fig. 60.

Produce CD to M; it is evident, that AL is to BM;

=

K