.. the points G, E, C, D are in the circumference of the same circle, i. e. the circle circumscribing ECD passes through G. Also since the angle GAE= GFD=GBD, i. e. GAC=GBD, .. the points G, A, C, B are in the circumference of the same circle; or the circle circumscribing ACB also passes through G.

(41.) Having given the base and vertical angle of a triangle; to determine the locus of the extremity of the line which always bisects the vertical angle, and is equal to half the sum of the sides containing the angle.

E

H

I

B

Let AB be the given base; and on it describe a segment of a circle ACB, containing an angle equal to the given vertical angle. Complete the circle; and draw the diameter FD bisecting AB. Join AF, FB; and with the centre F, and radius FA, describe a circle ABE, cutting FD in E. On DE as a diameter describe a circle; it will be the locus required.

Let ACB be any position of the triangle, and draw CGD; it bisects the angle at C, since ACD is equal to AFD, i. e. to half of AFB or to half ACB. Produce AC, AF to I and H. Join HI, CF, EG. The angle CAF is equal to CDF, and the angles AIH, FCD, DGE are right angles; .. the triangles AIH, CDF, EDG are equiangular,

and AH : AI :: FD : CD :: FE : CG. But AH is equal to AF and FB together, i. e. to 2 FE, and AI is equal to AC and CB together (ii. 60.), ... AC

and CB together are equal to 2 CG, i. e. CG is half the sum of the sides AC and CB, and its extremity is in the circumference of the circle EGD.

(42.) If from the extremities of the base of a triangle inscribed in a circle, perpendiculars be drawn to the opposite sides, intersecting a diameter which is perpendicular to the base; the segments of the diameter intercepted between these points and a point in it, whose distance from the base is equal to the lesser segment of the diameter made by the base, will be to one another in the ratio of the sides of the triangle.

Let ABC be a triangle inscribed in a circle, whose diameter DE is perpendicular to the base AC. Make FG = FE; and let the perpendiculars be drawn from A and C to the opposite sides, intersecting in H, and meeting the diameter in I and K;

=

B

GH

KG: IG :: AB : BC. Join AG, GC. Because GF FE, the angles GAF, GCF are each equal to FCE, i. e. to half the vertical angle of the triangle; .. AGC, ABC are together equal to two right angles (Eucl. i. 32); and since AHC is equal to its vertically opposite angle, AHC, ABC are equal to two right angles; whence AGC=AHC; and A, G, H, C are in the circumference of a circle; ... the angle GHA= GCA= half the angle ABC. Now the angle KHI, contained by the perpendiculars, is equal to ABC, .. GH bisects the angle KHI. Also the angle GKH=KHB=BAC; and KIH=AIF= ACB ;

.. the triangle KHI is equiangular to ABC; and it has been shewn that GH bisects the angle KHI,

.. KG

GI :: KH : HI :: AB: BC.

(43.) If the exterior angle of a triangle be bisected by a straight line which cuts the base produced; the square of the bisecting line is equal to the difference of the rectangles of the segments of the base and of the sides of the triangle.

Let CBD the exterior angle of the triangle ABC be bisected by BE which meets AC produced in E; the square of BE is equal

FB

D

to the difference of the rectangles AE, EC and AB, BC.

About the given triangle describe the circle ABC; and produce EB to F; and join AF. Then because the angle EBC= EBD = FBA, and AFB=BCE, since each of them together with ACB is equal to two right angles; the triangles EBC, FBA are equiangular, and AB BF :: EB : BC,

.. the rectangle AB, BC is equal to the rectangle EB, BF; to each of these equals add the square of BE, and the rectangle AB, BC together with the square of BE is equal to the rectangle EB, BF together with the square of BE, i. e. to the rectangle FE, EB, or its equal AE, EC; and.. the square of BE is equal to the difference between the rectangles AE, EC and AB, BC.

Squares

or Rectangles in connection

with a circle.

SECT. VIII.

(1.) IF from the centre of a circle a line be drawn to any point in the chord of an arc; the square of that line together with the rectangle contained by the segments of the chord will be equal to the square described on the radius.

From the centre O of the circle ABD, let OC be drawn to a point C in any chord AB; the square of OC together with the rectangle AC, CB is equal to the square described on the radius.

D

A

E

B

Through C draw DE perpendicular to OC. Join OD. Then DC=CE, and the rectangle DC, CE is equal to the square of DC; but the rectangle DC, CE is equal to the rectangle AC, CB, .. the rectangle AC, CB together with the square of CO is equal to the squares of DC, CO, i. e. to the square of DO.

(2.) If two straight lines in a circle cut each other at right angles; the sums of the squares of the two lines joining their extremities will be equal.

Let the two straight lines AC, BD cut each other at right angles in E; join AB, BC, CD, DA; the squares of AB, CD are equal to the squares of AD and CB.

For the squares of AB and CD are equal

B

D

E

to the squares of AE, EB, DE, EC. But the squares of AE and DE are equal to the square of AD, and the squares of EC and EB are equal to the square of BC; .. the squares of AB and CD are equal to the squares of AD and BC.

(3.) If two points be taken in the diameter of a circle, equidistant from the centre; the sum of the squares of two lines drawn from these points to any point in the circumference will always be the same.

D

А О В

E

Let A and B be two points in the diameter of the circle CDE, equally distant from the centre 0; if lines AC, BC be drawn to a point in the circumference, the sum of the squares of AC, CB will be the same, in whatever point of the circumference C is taken.

Join CO; then (iv. 30.) the sum of the squares of AC, CB is double of the sum of the squares of 40 and OC, which is an invariable quantity.

(4.) If from any point in the diameter of a semicircle there be drawn two straight lines to the circumference, one to its point of bisection, and the other at right angles to the diameter; the squares of these two lines are together double of the square of the semi-diameter.

From any point C in the diameter AB, let CD, CE be drawn; of which CD is pendicular to AB, and CE is drawn to the

per

B