(IR,IS) drawn from their point of concourse (1) parallel to the other two sides, (MO,LO),) trisect the buse.

Fig. 57.

Because IR is parallel to MO, the angle RIM is to IMO and .. to IMR, .. MR is to RI; for the same reason LS is = to 6F: again, because Ri is parallel to MO, the angle IRS is to OML, for the same reason ISR is to OLM, .. the triangle IRS is equilateral, .. IR and IS are each to RS, ... MR, RS, SL, are to one another,

PROP. 80, THEOR.

If from one extremity (A) of the base of an isosceles triangle (DAC) any right line (AK) he drawn to the opposite side, and from the same extremity an other right line (AB) be drawn to the same side produced, so that the intercept (KB) between the two drawn lines shall be equal to the first drawn line, the angle (KAD) contained by the first drawn line and the other side of the triangle is double the angle (CAB) contained by the second drawn line and base.

Fig. 31.

For the angle KCA is to CBA and CAB, ... it is = to KAB and CAB: also the angle DAC since it is to DCA, is to KAB and CAB, i. e to KAC and twice CAB,. ..if KAC be taken away, DAK will remain = to twice CAB.

PROP. 81. THEOR.

The sum of the sides of an isosceles triangle (ABC) is less than the sum of the sides of any other triangle (DBC) on the same base (BC) and between the same parallels.

Fig. 49.

For, since FD is parallel to BC, the angles FAB,ABC are, also the angles DAC,ABC,.. the angles DAC, FAB are, .. the sum of the right lines BA and CA

is less than the sum of any other two lines BD and CD that can be drawn from the points B, C to FD (prop. 10 of deducibles.).

PROP. 82, THEOR.

If the exterior angles (OKL,KLR) of any triangle (KML) be bisected by the right lines, (KN,LN,) the right line (NM) drawn from the point of concourse (N) to the opposite angle bisects it.

Suppose the angles OMN,RMN are, draw the perpendiculars NR,NO: they are evidently, (by prop. 26. 1. Elr.) if this can be proved the proposition is true.

Draw N1 at right angles to KL. Because the angles NLR,NLI are, and also the angles NIL,NRL and the site NL common to the triangles NIL,NRL, NI is to NR, for the same reason N1 is = to NO, .. NO is to NR. Then since NR is to NO and NM common to the triangles NMR,NMO, and the angles NRM,NOM right angles, the angies NMR, NMO must be of the same affection,.. the triangles are in every respect, and.. the angles NMR,NMO are =.

PROP. 83. PROB.

To draw from the extremity (A) of one side [AB] of a parallelogram to the opposite side, (KB,) a right ine which shall be equal to the side to which it is drawn, together with the segment of it intercepted between the drawn line and adjacent angle [K.]

Suppose it done, and that AP is to the sum of BK and KP. Produce PK till PO is to PA, join OA. Because PO is to PA, PO is to BK and KP, KO is to KB, also the angles POA,PAO are =.

Therefore it is evident, if you make KO≈ to KB, join OA, and make the angle OAP to POA, that AP is the line required to be drawn.

PROP. 84, THEOR.

The area of a trapezium (LS) is half that of a parallelogram (LB) whose base (AB) is equal to the sum of the parallel sides (LC,AS) of the trapezium, and whose altitude is equal to that of the trapezium.

Fig. 59.

Draw CT, RS parallel to LA or KB.

Then, since LE is to SB, AS is to TB,.. the parallelograms RA and ETS is to SRE, .. ESBK,.. &c. &c.

EB are, also the triangle the trapezium LASE is to

PROP. 85, THEOR.

If perpendiculars (PI, CI, DI) be drawn from the middle points of the sides of an equilateral triangle 'till they meet in 1, they are équal to one another; and their sum is equal to the perpendicular [MC] drawn from any angle (M) to the opposite side,

Fig. 57.

It is evident, that if one of the perpendiculars CI be produced, it will bisect the angle OML: then, because DM is to MP and MI common and the angles DMI, PMI =3 Pl is to ID, for the same reason Pi is = to IC,.. they are to one another.

And because CM bisects the angle OML, and also DL bisects the angle OLM, they cut one another in a point of trisection, .. CI is the third part of CM, but I is also the third part of the sum of the three, DI, PI, aud CI,.. the three together are to CM.

PROP. 86, THEOR.

In proposition 52, the internal square is half the external, when the sides of the external square are bisected.

Fig. 44.

Draw a diagonal BC: then because the sides of the triangle ACB are bisected in a and b, the triangle A ab is the fourth part of, ABC, and ... the eighth part of the AI, for the same reason each of the triangles a C d, d le, c B d, is the eighth part of the entire figure,.. the four together are to half AI, .. the square abcd is half the square AI.

PROP. 87. PROB.

Given of any triangle the base, the sum and difference of the angles at the base to construct it.

Suppose ABD to be the required triangle, cut from AD a part AC to AB, join BC.

Then, because AB is to AC, it is evident (from prop. 18 of dedu.) that the angle CBD is to half the difference of the base angles, and also that the angle ABC is to half the sum, .. ABD is to half the sum + half the difference, and ... ADB is to half the sum half differenee.

From this it is evident, if from one extremity B, of the given base, you draw BA, making the angle DBA = to half the sum + half the difference, and if from the other extremity D you draw DA, making the angle BDA = to half the sum half the difference, that ABD is the required triangle.

PROP. 88. PROB.

Given in any triangle, one side, an angle at it, and the length of the perpendicular drawn from its middle point, to meet those drawn from the middle points of the other two sides, to construct the triangle.

Suppose GAC to be the required triangle. Let GC be the given side, CGA the given angle, and EI the

given perpendicular; join IG,IA, bisect GA in B, and draw BI, it is at right angles to GA, (prop. 32 of dedu.) IA is to IG.

Then at one extremity, G, of the given side, make the angle CGA to the given angle, draw El to the given perpendicular, join IG, and from I draw a line IA to IG to meet GĂ, join AC; AGC is evidently the required triangle.

PROP. 89. PROB.

Given of any triangle, one side and the lines which drawn from the angles at it, bisect the other two sides, to construet it.

Suppose DCE is the required triangle. Let CE be the given side, and CF,EA the given bisectors; it is evident (prop. 24 of dedu.) that they cut one another in a point of trisection, . CI and El are respectively 3rds of CF and EA.

If therefore, with the given side CE and with CI, EI the respective 3rd parts of the given bisectors, you form a triangle CIE, produce those parts untill CF and EA are respectively to the given bisectors, draw through their extremities A,F, (from the extremities of the given side,) CD and ED, DUE is the required triangle.

PROP. 90. PROB.

Given of any triangle, one side, an angle at it, and one of the equal perpendiculars on the sides, to construct the triangle.

Suppose ABC to be the required triangle. Let BC be the given side, BCA the given angle, and DF one of the perpendiculars; join DC, and draw the perpendicular DG. It is evident (prop. of dedu.) that DC

=

bisects the augle BCA.

If therefore, you bisect the given angle by CD and subtend the augle BCD with DF to the given perpen