Fig. 19.

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

PROP. 103. PROB.

in

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,IBE are, and because IH is parallel to BA, the angles IEB,EBA are,.. the anglés IEE,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; then through M draw MK parallel to FE; the sum of KE,MF is evidently to that of PG,OH: through L draw LI parallel to PK: IK is = to LP, .. the sum of IK, KE, BM and MF, (of one pair of perpendiculars CE, BF,) is to the sum of AL, LP, PG and OH, (of the other pair,) if.. CI is

to DO, the pro

position is true. Because IC is parallel to OD, the angles ICL, ODM are, for the same reason the angles CIL, DOM are, and CL is evidently to DM, .. CI is to DO, (prop. 26 1 Elr.) .'. &c. &c.

PROP. 105, THEOR.

=

In an equilateral triangle, (EFG,) the perpendiculars [RP,RS] let fall from any point (R) in one side, are together equal to the perpendicular, (FO,) from any argle on its opposite side.

Fig. 61.

Draw RL parallel to GE; then XO is evidently to one of the perpendiculars RS, if then PR is to FX the proposition is true. Because LR is parallel to EG, the angles FXR,FOG are =,.. FXR,FPR are =, for the same reason the angles XRF,PFR are =, and FR common to the triangles PFR,XFR, .. FX is = to RP, (prop. 26. 1. Elr.).. &c. &c.

PROP. 106. THEOR.

If points (D,F,H) be taken in the side (AB) of any triangle, at equal distances from one another, and lines (DE, FG and HI) be drawn through them parallel to the base, the parts (EG, GI) of the other side, intercepted between them, are also equal.

Fig. 62.

Draw through G, OGL parallel to BA, and produce DE to meet it.

Then the right lines LG, GO are evidently = to one another, and since DL is parallel to HF, the angles ELO, LOF are, for the same reason the angles LEF, EFO are, .. EG is = to GF.

Cor. From this, it is evident, that if one side of a triangle be divided into any number of equal parts, and if right lines be drawn from the points of section parallel to another side, they divide the third side into the same number of equal parts,

PROP. 107. PROB.

To draw from the side [AB] of any triangle [ABC] to the adjacent side [BC] a right line equal to the upper segment of the side from which it is drawn, and making with the side to which it is drawn, an angle equal to a given rectilineal angle.

Suppose EF is the line required to be drawn; join FA, and draw AG parallel to FF: the angle AGC is = to the given angle, draw through the angle AGC the right line GD parallel to FA to meet BA produced. Then, because GD is parallel to FA, the angles DGA,FAG are, for the same reason the angles EFA,FAG are ,.. the angles AGD,EFA are, also the angle EAF (EFA) is to ADG, . AD is =

=

=

to AG. Then, through A draw AG, making the angle AGC to the given angle, produce BA until AD is to AG, join DG; through A draw AF parallel to DG, and through F draw FE parallel to GA: FE is the line required to be drawn.

- =

For the angles EFA, FAG, AGD and ADG are evidently to one another, and EAF is to ADG .. to EFA.. EF is to EA, and the angle EFC is to AGC, and ... to the given angle.

=

Cor. In the same way a line can be drawn from à side of a triangle to the adjacent side, equal to the upper segment, making an angle with the side from which it is drawn, equal to a given rectilineal angle.

PROP. 108. THEOR.

If the altitude of any parallelogram (HV) inscribed in a triangle, (ABC,) is half the altitude of the triangle, its base shall be equal to half the base of the triangle.

Fig. 64,

For, draw AI at right angles to the base BC.

Then, because AI is bisected in U, and UV parallel to BC, AC is also bisected in V, for the same reason CI is bisected in K, and also BI in H, &c. &c.

PROP. 109, THEOR.

Of all the rectangles that can be inscribed in a triangle, (ABC,) the greatest is that (LK) whose sides are respectively equal to the semi-base and semi-altitude of the triangle.

Fig. 64

For, let DG be any other parallelogram, it is less than LK. Draw AI at right angles to BC, join UH, DH and OH.

.'.

LU is evidently and parallel to BH, .. UH is = and parallel to LB, .. the parallelograms LI and LH are to one another; and since HU is parallel to BA the triangles LHU,DHU are, but LHU is half of the parallelogram LI, .. DHU is to half of it; also the triangle OHU is half of the parallelogram OI, and the triangle ODU is half of DU; but both of these triangles are together less than DHU, .. their doubles or DI is less than double DHU or LI: for. the same reason El is less than UI, .. DG is less than LK.

Cor. It is evident from this, that the greatest parallelogram that can be inscribed in a triangle, is that whose base is the semi-base, and altitude the semi-altitude of the triangle. For, draw UH parallel. to AB, it is evident that the parallelogram UB is to LK, .: &c. &c.

PROP. 110, PROṚ.

To inscribe a square in any triangle (ABC).

Suppose that DG is the required '; draw the dia gonals FE,GD: it is evident that they bisect one another, and also that each of the angles OFG,OGF is half a right angle; draw any lines HL,IK respectively parallel to FE,GD, the angles LHC,KIB are each to half a right angle; also draw OB and OC, since those lines bisect the diagonals they bisect the lines parallel to them, (prop. 14, dedu.)

Then it is evident, if you draw any lines HL,IK, making each of the angles KIB,LHC to half a right angle, bisect those lines and through the points of bi

K

section draw BO and CO, the lines EF,DG drawn thro' the point in which they meet respectively parallel to LG,KI, are evidently the diagonals, the required 2: for, draw GE,ED and DF:

Then, because HL is bisected by CO, EF is also bisected by it, (prop. 14, dedu ) for the same reason DG is bisected by BO; and because DH is parallel to KI, the angle DGF is half a right angle, also EFG is half a right angle, .. FOG is a right angle, ... each of the angles FOD, DOE, EOG is a right angle, and since FO is evidently to GO, the four GO, EO, DO and FO are to one another, .. D F G E is a square.

PROP. 111, THEOR.

If perpendiculars be drawn from three angles of a parallelogram (AD) to a right line drawn from the fourth angle, the sum of those (CE,AF) from the remote angles, is equal to that (DL) from the adjacent angle.

Fig. 66.

=

Draw EG parallel to CD, GD is evidently to EC, and since EG is = and par. to CD, it is also and par. to AB, .. in the triangles ABF,LEG the angles ABF, BFA are respectively to LEG,ELG, and the side EG to AB, ... AF is to LG, .. AF and EC together are to LD.

PROP. 112, THEOR.

To draw from one side (AB) of any triangle to the adjacent side, a right line equal to half the upper seyment of the side from which it is drawn, and making with this side an angle equal to a given rectilineal angle.

Suppose that GH is the line required to be drawn: produce it 'till GF is to GA, join AF; the angles FAG,AFG are, through B draw BD parallel to GF and produce AH,AF to meet it, BD is evidently to BA and is bisected in E, and the angle ABD is to AGF, and to the given angle.