B

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.

=

D

E

Join EC. Since ABAC, 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 drawn 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

E

B draw AD, BD to any point D in the circumference, cutting the base CG in E and F;

CE CF:: CF: CG.

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,

.. ex æquo BF

CG: AB: BD :: EF: FD,

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.

On AB, BC, the sides of the triangle ABC, let the segments ADB, BEC be described, similar to AFC the segment on AC. At A and C let tangents AD, CE be drawn. Join DB, BE; they are in the same straight line.

D

H

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 O; O is the centre of a circle which will touch both the semicircles.

From O draw OFE, OHD perpendicular to the sides.

OH being parallel to BC,

Then

(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

D and E, let circles be described touching that described on the base in F and G; each of these circles will touch the semicircle described on the other side.

Join ODF, OEG, DE. Since AB, BC are bisected in D and E,

DE is parallel to AC, and equal to half AC, i. e. to AO or OC. In the same manner OD is parallel to BC, and OE to AB; .. ODBE is a parallelogram, and EB, i. e. EH=OD; but OF=DE, .. DH= DF, and H is a point in the circumference of the circle FHK; and being in the circumference of BHC, it will be the point of contact, since DE joins the centres. In the same manner it may be shewn that the circle GI touches the circle ABI in I.

(21.) If from any point in the circumference of a circle perpendiculars be drawn to the sides of the inscribed triangle; the three points of intersection will be in the same straight line.

From D any point in the circumference

of the circle ABC, let DE, DF, DG be drawn perpendicular to the sides of the inscribed triangle ACB; join EF, FG; EFG is a straight line.

A E

Join AD, BD, CD. Since the angles DFB, DGB are right angles, a circle may be described about the quadrilateral figure DGBF (vi. 13.); and the angle DFG is equal to DBG. Also since the angles DFA, DEA are right angles, a circle may be described

about the quadrilateral figure DFEA; whence the angles DFE, DAE are together equal to two right angles. But ADBC being a quadrilateral figure inscribed in a circle, the angle DAC is equal to DBG, i. e. to DFG ; .. DFE, DFG are equal to two right angles; and EFG is a straight line.

(22.) The base of a right-angled triangle not being greater than the perpendicular; if on any line drawn from the vertex to the base a semicircle be described, and a chord equal to the perpendicular placed in it, and bisected; the point of bisection will always fall within the triangle.

B

F

H

G

Let ABC be a right-angled triangle, of which the side AC is not greater than BC. From B let any line BD be drawn to the base; on which describe a semicircle BCD, and in it place EF= BC, which bisect in G; the point G is within the triangle ABC.

per

Since BC is equal to EF, angles at G and H being

Take O the centre of the semicircle; draw OH pendicular to BC; join OG. OH is equal to OG; and the right angles, a circle described with the centre O, and radius OG, will touch BC in H, and .. G is within the angle D BC. Also since AC is not greater than BC, DC is less than BC or EF, .. EF is nearer to the centre O, than DC is; or G falls above DC and within the angle DCB.