dicular, and making with CB the angle CFD a right angle; (prop. 27, dedu.) join DB and draw BA, making the angle DBA to DBC; ABC is evidently the required triangle.

PROP. 91. THEOR.

The square of the hypothenuse (AC) of an isosceles right angled triangle (AEC) is equal to four times the square of the perpendicular (ED) from the right angle on the hypothenuse.

Fig. 19.

For ED is to AD or DC: but the squares of AE, EC are to two squares ED with the squares of AD and DC, but the squares of AD and DC are = to two squares of ED,.. the rs of AE,EC are = to four []'rs of AC, which is the 'rs of 'rs of ED.

of ED, and .. the □ AE, EC is to four

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 IE, IF, IB at right angles to the middle points of the sides; join 1A and IG. 1A 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, 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 the sum of the hypothenuse and one side, and the contained angle of a right angled triangle, 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 ČD; and the angle DAC is the complement of the given angle, because it with DCA is 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 and CDB, and .. to the given angle; DC and CA are to AB, because the angles BDC,CBD are, and the angle CDA is a right angle, because the angies CAD,ACD are 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 BA and AK, .. than BA and AC; .. half BK or AD is less than half CA and AB.

PROP. 95, PROB.

Given one side and the perpendicular on the hypothe nuse 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 and BAO are to a right angle,.. OBC and 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,CGF, the angles ICF and GCF are, also the angles CIF,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 opposite angles, (FCA,AEF,) to the points of bisection (D,G) of their opposite sides, (AE,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 A1,IK and KF are to one another, .. the diagonal is trisected.

=

=

PROP. 98, THEOR.

If the right lines (DF,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.

Becanse in the triangles FAD,GCD the sides FD and DA of the oue are respectively = to GD and DC of the other; the angles DFA,DGC =, and the angles FDA, GDC of the same affection, (since the triangles are right angled,) the angles GCD,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 of the point [F] where it cuts the greater side, from 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, .. A is half the sum of BA and AD. And since BA is to AC, CD is the difference between BA and AP, ... FD is half the difference.

PROP. 100. PROB.

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

Suppose that ABD is the required triangle. Let BD be the given 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 —, .'. 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 hypothenuse and side, to construct it.

Fig. 37.

ANALYSIS.

Same as 43

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,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,DG] let fall from the middle point of the base of an isosceles triangle on the sides are equal.

+