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PROP. 54. THEOR.

If from a given point [O] within a circle which is not the centre, right lines [OC OD OA] be drawn to the circumference, making equal angles with each other, those [OD,DA] which are next that part of the diameter [OL] passing through the given point, shall cut off a greater circumference than the two which are more remote.

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Suppose that the circumference DA is greater than CD, join CD DA; then the right line DA is greater than CD, draw OL through the centre R, then it is evident that OC is less than DA, make 01 to OC join DI then it is evident the triangles OCD OID are

in every way,.. DI is to DC, ... DA is greater than DI, if then this is proved the proposition is true! Produce DO and AO to S and P, and join AS AC and CS.

Then the angle AID is = to the angles IOD, IDO= to the angles DOC ODC, and the angle DOC is greater the angle OSC, and OSC is = to DAC, [prop.21.b.3. Elr.] .. DOC or DOI is greater than DAC, also the angle CDS is to the angle CAS,.. CDO and COD IDO IOD are together greater than DAS, ... DIA is greater than DAS and .. than DAI... DA is greater than D1, but DI is to DC .. DA is greater than DC.

PROP. 55. PROB.

Through a given point[A],within a circle, which is not the centre, to draw the least possible chord.

Fig. 57.

Find the centre K, join KA and through A draw CP at right angles to KA it is less than any other chord DL. For, draw KS at right angles to DL: then since the angle KSA is right, KA is greater than KS.. CP is less than DL, in like manner it can be proved, that CO is less than any other right line drawn through A, and terminated by the circumference.

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PROP. 56. PROB.

Through either of the points of intersection [A] of two given circles, which cut each other, to draw the greatest possible line, which shall be terminated both ways by the two circumferences.

Fig. 58.

Join the centres O,S, and draw through A the right line CD, parallel to OS it is greater than any other right line, BF drawn through A.

For draw OL SI at right angles to CD, then LI is evidently half of CD; also draw OP SQ at right angles to BF and draw ST parallel to FB: it is also evident that ST is half of BF, and because ST is parallel to BF, STO is a right angle, .. SO is greater than ST, .. LI is greater than PQ, and .. CD is greater than BF, in like manner CD can be proved greater than any other right line drawn through A and terminated by the circumferences.

PROP. 57. THEOR.

Of all the triangles [AEC AFC] standing on the same base [AC] and having equal vertical angles, the greatest is that [AEC] which is isosecles.

Fig. 59.

Let A EFC be a segment of a circle in which the = angles AEC,A FC are contained, draw EL perpendicular and FI par. to AC, from the centre O draw OF and join IA and IC.

Then it is evident that EL bisects AC, and also passes through the centre, .. OF or OE is greater OI, .. the triangle AEC is greater than AIC, but AIC is to AFC, .. AEC is greater than AFC, ... &c.

PROP. 58. THEOR.

Of all the right lines [AF EC] that can be drawn to cut off equal areas, from a given triangl [IDL] that [AF] is the least which makes the triangle [r] cut op an isocecles one.

Fig. 60.

Let ICH be a circle passing through the points I, E, C; bisect ECin O, join EH CH and OH.

Then the triangle ECH is isosceles and is greater than IEC (by the feregoing prop.) and is .. greater than IAF: then since the triangles HEC IAF are equiangular, and IAF less than EHC the right line AF must consequently the greater than EC.

PROP. 59. THEOR.

If through the centre [C] of a given circle another [LCI] be described with any radius, the intercepts [EF FD] of any secant. [AD], from the remote term [A] of a right line [AB] joining their centres and produced to meet the circumferences, are equal. Fig. 61.

For join CE CF and CD; then the angles CED CDE are (prop.5.b.1.Elr.) and the angle CFE, being an angle in a semicircle, is right ... CFD is right; then in the triangles CFE CFD the angles CEF CFE of one are respectively to CDF CFD of the other and CE= to CD .. EF is DF (prop. 26. b. 1. Elr.) ̧.°.

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PROP. 59. THEOR.

If from a side [AC] of a given rectangle, a part [AI] be cut of equal to half another side [AE] and from the point of section, [I] a right line [IL] be inflected equal to the cut off part [IA], the square of the right line [LA] joining their extremities is equal to the given rectangle: and also the square of the right line [EO] drawn from the opposite angle at right angles to the connecting line, [AL] is equal to the given rectangle.

Fig. 62.

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1. For the of AL is to the sum of thers of AI IL with two rectangles under AI IC; but the 'rs

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2. Also the of OE is to the rectangle CE. For bisect AE is G join GO and let fall the perpendicular DF: then GO is to GE IL, and the angle FEO = AOF = IAL = ILA is to GOE; also the angle CLA = LAE is to FOE.. the angle CLI is to FOG and LCI to OFG and OG to IL .. OF is = to LC: and since in the triangles LCA OFE the angles LAC ACL of the one are respectively to OEF EFO LC, OE is to LA but LA iş

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PROP. 61. PROB.

In a given right line [EF] to find a point at which right lines drawn from two given points[A,H]without, it shall contain an angle which shall be a maximum. Fig. 63.

Join AH if it be parallel to EF, bisect it in C and draw CD at right angles to EF, join AD HD, and the angle ADH shall be a maximum: for describe a circle to pass through AHD; EF is evidently a tangent to it, and then the angle ADH is evidently greater than any other angle AIH: for join AO, and the angle AOH is to the angle ADH but it is greater than AIH,.`. &c.

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But if AH is not par. to EF, produce them till they meet, find the side of a2 to the reet. HEA, make ED = to it, and through AHD describe a circle, ED is evidently a tangent to it, join AD HD the angle ADH is a maximum. This may be proved as the foregoing

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PROP. 62. THEOR.

Let ADBC be a given circle, APC a given triangle inscribed in it, DP a diameter bisecting AC, and PI a perpendicular on AB, then IB shall be equal to half AB BC.

Fig. 64.

For draw PX at right angles to BC produced and join BP; then(prop.26.b.1. Elr.)BI is to BX and PI to PX; also AP is to PC.. AI is to CX and ... IB, BX are together to AB, BC together, but1B is to balf the former quantity, and .. to half the latter,

Cor. 1. Hence IB BC= AI and BA =BC + 2 AI .. AI is to half the difference of AB, BC.

Cor. 2. Let fall DS at right angles to AB, then AS is to IB; for draw BH parallel to AC and join AH, which, on account of the parrallels is — to BC, but BS by the proposition is to half the difference of BA AH or BA BC ... AI and AS IB, `

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PROP 63. PROB.

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

Fig. 65.

Describe on the given base AB a segment of a circle capable of containing the given angle, complete the the circle and draw the diameter EF at right angles to the base, join AF on AF describe a semicircle, and in it place AO half the given difference, produce AO to I and join IC, IAC is the required triangle.

For join FO, then (by the foregoing prop.) AO is= to the difference between OI, and IC and.. 2 AO is to the difference between AI IC.

PROP. 64. THEOR,

Given the sum of the sides, the difference of the segments of the base, and difference of the base angles to find the triangle.

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Let XIO be the triangle, make XD to the sum of the sides, join DO; then evidently the angle XOD is given to a right angle + half the difference, and if XY difference of the segments of the base, and NX the difference of the sides, DX X XN = 0X X XY.. DNYO could be inscribed in a circle .. the angle DNY is given,

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