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circle, and through the angular points another be circumscribed; to determine the ratio which they bear to each other.

Let ABC be an equilateral triangle inscribed in the circle, about which another DEF is circumscribed, touching the circle in the points A, B, C.

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Since DA touches the circle, the angle DAB ACB (Eucl. iii. 32.); but ACB=ABC; :. DAB=ABC, and they are alternate angles, :. DF is parallel to BC. In the same manner it may be shewn that AB is parallel to FE, .. ABCF is a parallelogram ; and the triangle ABC is equal to AFC. In the same manner ABC may be shewn to be equal to each of the triangles ABD, BCE; and .. it is one fourth of the circumscribing triangle.

(14.) A straight line drawn from the vertex of equilateral triangle inscribed in a circle to any point in the opposite circumference, is equal to the two lines together, which are drawn from the extremities of the base to the same point.

Let ABC be an equilateral triangle inscribed in a circle; from B draw BD to any point D in the circumference. Join AD, CD. BD is equal to AD and CD together.

Make DEDA, and join AE. The

angle DAE is equal to the angle DEA; but ADE = ACB in the same segment, .. DAE and DEA together are equal to CBA and CAB together; whence DAE

=CAB; and taking away the common angle CAE, DAC = EAB; but DCA EBA, and AC=AB, :. BE= DC; and BD is equal to AD and CD together.

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(15.) If the base of a triangle be produced both ways so that each part produced may be equal to the adjacent side, and through the extremities of the parts produced and the vertex a circle be described; the line joining its centre and the vertex of the triangle will bisect the angle at the vertex.

Let AC a side of the triangle ABC be produced both ways till AD =AB, and CE=CB; and through D, B, E let a circle be described,

whose centre is O. If OB be joined, it will bisect the angle ABC.

Join BD, BE, OA, OD, OE. Since DA=AB, the angle ABD is equal to ADB; but the angle OBD is equal to ODB, and .. the angle OBA is equal to ODA. In the same manner it may be shewn that the angle OBC is equal to OEC; and since ODA is equal to OEC, OBA is equal to OBC; or ABC is bisected by OB.

(16.) If an isosceles triangle be inscribed in a circle, and from the vertical angle a line be drawn meeting the circumference and the base; either equal side is a mean proportional between the segments of the line thus drawn.

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Let ABC be an isosceles triangle inscribed in the circle AEC, the side AB being equal to AC; and from A draw any line AED, meeting the circumference in E, and CB produced in D; AB is a mean proportional between DA and AE.

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Join EC. Since AB= AC, the angle ACB=ABC = AEC in the same segment; and the angle at A is common to the triangles AEC, ACD, .. the triangles are equiangular and similar;

. AD: AC :: AC: AE.

(17.) If from the extremities of one of the equal sides of an isosceles triangle inscribed in a circle, tangents be drawn to the circle, and produced to meet; two lines draum to any point in the circumference from the point of concourse and one point of contact will divide the base (produced if necessary) in geometrical proportion.

Let CBG be an isosceles triangle inscribed in a circle, the side CB being equal to BG; and at B and Clet tangents BA, CA be drawn, meeting in A. From A and

B draw AD, BD to

E

F

any point D in the circumference, cutting the base CG in E and F;

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Join CD. The angle ABC being equal to BGC, is equal to BCG, and .. CG is parallel to AB; and the triangles ABC, CBG are equiangular;

.. BC CG :: AB; BC;

but (vii. 16.) BF : BC :: BC : BD,

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CG AB: BD :: EF: FD,
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since CG is parallel to AB;

but CF

BF :: FD : FG,

.. CF CG :: EF: FG,

whence (Eucl. v. 19.) CE CF CF CG.

(18.) If on the sides of a triangle, segments of circles be described similar to a segment on the base, and from the extremities of the base tangents be drawn intersecting their circumferences; the points of intersection and the vertex of the triangle will be in the same straight line.

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Since DA touches the circle AFG, the angle DAC is equal to the angle in the alternate segment AGC, i. e. to the angle in the segment AHB. But the angle ADB, together with the angle in the segment AHB, will be equal to two right angles; .. the angles CAD, ADB are equal to two right angles; .. AC, DB are parallel. In the same manner AC, BE may be shewn to be parallel; .. BD and BE being drawn from the same point, parallel to the same line, will also be in the same straight line.

(19.) The centre of the circle, which touches the semicircles described on the two sides of a right-angled triangle, is in the middle point of the hypothenuse.

On the sides AB, BC of the right-angled triangle ABC, let semicircles ADB, BEC be described. Bisect AC in 0; 0 is the centre of a circle which will touch both the semicircles.

From O draw OFE, OHD perpendicular to the sides. Then OH being parallel to BC,

(Eucl. vi. 2.) AO: OC :: AH : HB,

.. AH=HB, and H is the centre of the semicircle ADB. Hence the centre of a circle touching ADB in D is in the line DHO. For the same reason, the centre of a circle touching BEC in E is in the line EFO. Also since OD is equal to OH and HD together, i. e. to BF and HB, or EF and FO together, i. e. to EO, O is the centre of the circle, which will touch both.

COR. The diameter of this circle will be equal to the sides of the triangle together.

(20.) If on the three sides of a right-angled triangle semicircles be described, and with the centres of those described on the sides, circles be described touching that described on the base; they will also touch the other semicircles.

On the sides AB, BC of the right-angled triangle ABC let semicircles be described; and with the centres

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