Suppose that I is the point required to be found; that CH is at right angles to AB and that C1 is to IA; then, because CI and IA are, the angles IAC,¡CA are ; draw AX parallel to CH: because CH is parallel to XÁ the angles CAX,ACI are =,.. CAX,CAD are , and also the angle XAB is a right angle.

If therefore you draw AX at right angles to BA, bisect the angle DAX, draw the right line AC bisecting it to meet BD produced, and through C draw CH parallel to XA; the point I where it cuts AD is the required point.

For, since HC is parallel to AX, the angles HCA,CAX are,.. ICA, IAC are =, .. IC is to IA; and the angle CHA is evidently a right angle, .. &c. &c.

PROP. 135. THEOR.

If the base [BD] of a right angled triangle [BDG] be divided into any number of equal parts (DE, EF, FB,] and right lines be drawn from the points of section to the opposite angle [BGD] the angle [EGD] subtended by the segment [ED] adjacent to the right angle is greater than either of the remaining ones, and the segment nearest to this shall subtend a greater angle than the more remote.

Fig. 23.

For, since the angle BDG is right, the angle DBG is less than it,.. (prop. 19, dedu.) if the angles EGD,EGF were EF would be greater than ED, a fortiore if the angle EGF was greater than EGD, EF would be much greater than ED but it is to it,.. the angle FGE must be less than EGD; for the same reason the angle BGF is less than FGE.

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PROP. 136, PROB.

In a given square [DC] to inscribe an equilateral tri ingle, one of whose angular points shall be in an an angular point [B] of the given square.

Fig. 11.

Draw a diagonal BE; at the point B and with the right line BE make the angle EBP = to half the angle of an equilateral triangle, make the angle EBI = to EBP, join IP: IBP is the required triangle.

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For in the triangles DBP,IBC the side DB is = to the side BC, and the angles BCI,CBI of the one are respectively to BDP,PBD of the other, .. BP is to BI, the angles BIP,BPI are, but they are together 120 degrees, .. each of them is 60, .. the triangle BPI is equilateral.

PROP. 137. PROB.

To find a point in the side [BD] produced, of any triangle, from which a right line drawn at right angles to the other side [BA] and cutting the base [AD,] may be equal to the base.

Suppose that HC is the line required to be drawn: through A draw AO and parallel to HC, through O draw OC; OC is and parallel to AH.

If therefore through the extremity A of the side to which the line is to be drawn, you draw AO at right angles to it and to the base, through O draw OC parallel to ABC is the point required to be found. For draw CH parallel to AO, it is evidently to it and is .•. = to the base AD, it is also at right angles to AB, since it is parallel to BO and OAB being a right angle,

PROP. 138. THEOR.

If any two points [e F] be taken in the opposite sides ofaparallelogram [CEHG] at equal distances from the opposite angles, [ECG,EHG,] and if two other points [k, o] be taken in the other two sides at equal distances from the same angles, the right lines [eo, oF, Fk, ke] connecting those points form a parallelogram. Fig. 51.

For in the triangles k Ce, o HF the sides k C, Ce of the one are respectively to o H, HF of the other,

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and the contained angles are =,.. ek is to Fo, and because Ce is to HF, e G is to FE, for the same reason E k is = to o G, and the contained angles o G e, FEk,..k F is to eo,.. eo Fk is a parallelogram.

PROP. 139, THEOR.

If one of the diagonals (CB) of a given parallelogram (DBEC) be equal to a side (DB or CE) of the parallelogram, the other diagonal (DE) shall be greater than any side of the figure.

Fig. 15.

Because CB is to CE the angles CBE, CEB are = to one another, .. the angle CBE is greater than the angle DEB, much greater then is the angle DBE than DEB; for the same reason the angle DBE is much greater than the angle EDB,.. DE is greater than either of the sides DB or BE and .. than either of the sides DC or CE.

PROP. 140, PROB.

To describe a circle through three given points (A, C, G) that are not in directum.

Suppose it done; connect the given points by the right lines GA, AC, CG, and also the centre of the circle I and the given points by IA, IG, IC, those lines are to one another, (def. 15 & 16,) let fall from 1 on the sides AC, CG, GA, the perpendiculars IB, IE and IF; then in the triangles AIF,CIF the side IA is to IC, and the angles IAF,IFA of the one respectively = to ICF,IFC of the other, .. (prop. 26, 1 Elr.) AF is to FC, for the same reason CE is to EG and GB to BA.

If therefore you bisect the lines connecting the given points, draw perpendiculars from the points of bisection to meet; the point in which they meet is evidently the centre of the circle required to be described.

PROP. 141. PROB

Given the base (AC) to find the locus of the vertex of an infinite number of triangles, the differences of the squares of whose sides shall be equal.

Suppose ABC and AEC to be any number of triangles the difference of the squares of whose sides are≈; then (by cor. 4, prop. 48, 1 Elr) the difference between the squares of AB,BC is to the difference between the 'rs of the segments AD,DC of the base made by the perpendicular BD from the vertex on the base, also the difference between the sides AE,EC is to the difference between the 'rs of the segments of the base, but since the differences are the segments must be =, .. the locus of the vertex lies in a line drawn at right angles to the base.

PROP. 142. PROB.

Given any triangle (BAI) one of whose sides (BI) is produced, to let fall from the produced side, on the other (AB,) a perpendicular, which shall be bisected by the base (BI).

Suppose that GF is the line required to be drawn ; draw through B the right line BD parallel to GF and to meet the produced side; draw AH and produce it to E; since GF is bisected by AE, BD is also bisected by it, (prop. 14, deduc.)

Therefore let BD be drawn to meet AF produced and at right angles to AB, bisect BD in E, join EA; it is evident the right line FG drawn through H parallel to BD is at right angles to AB and bisected by BI.

PROP. 143. THEOR.

The right line (DA) drawn from the point of concourse D, of two lines (DB,DC) bisecting any two angles of a triang e (ABC) to the remaining angle (BAC) bisects it also.

Fig. 50.

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Draw DE, DF and DG at right angles to the sides; they are evidently to one another; and because in the triangles AED, AGD the side AD is common, the angles AED, AGD right angles, and the angles GAD,EAD of the same affection, the triangles are in every way, (prop. 77, dedu.) .. the angles GAD,EAD are =, ... &c.

PROP. 144, THEOR.

That point within a triangle [MOL] from which lines [IO, IL, IM] drawn to the oppoite angles, trisect the triangle, is the occourse of three lines bisecting the sides.

Fig. 57.

Because ML is bisected in P the triangle MPO is half of MLO, for the same reason the triangle DML is half MLO,.. DML,MPO are; if then the figure MPID be taken from both the triangles, PIL,DIO will remain to one another; and because ML is bisected in P the triangles PIM,PIL are,.. PIM,DIO are =, for the same reason DIM,PIL are =,. the triangle MIO is to MIL, in like manner the triangle LIO can be proved to LIM, .. the three triangles LIM, MIO, OIL are,.. the point of occourse of those lines is the point from which the lines drawn to the angles bisect the triangle.

PROP. 145. PROB.

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

Suppose CAB to be the required triangle: CB the given base, CK the given sum and CAB the given angle, draw KB, it is evident that AK is

AB and

the

angles AKB,ABK are, but the angle CAB is to AKB and ABK, ... AKB is to half the given angle.

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