(30.) To find the locus of the vertex of a triangle, whose base and ratio of the other two sides are given.

Let AB be the given base; divide it in C so that AC: CB may be in the given ratio of the sides.

Produce

AB to 0; and take CO a mean pro

B

portional between AO and BO. With the centre C, and radius CO, describe a circle; it will be the locus required.

In the arc AD take any point D; join DA, DB, DC, DO. Since OD= OC,

.. the sides about the common angle O are proportional, and the triangles ADO, BDO are equiangular; :. AD: DB :: DO : OB :: CO : OB :: AO : CO :: AO-CO: CO- OB :: AC : CB,

i. e. in the given ratio. In the same manner, if any other point be taken in the circumference of the circle, and lines drawn to it, they will be in the same given ratio, and .. the circumference is the locus required.

COR. Since in any triangle, if from the vertex a line be drawn cutting the base in the ratio of the sides, it will bisect the angle, .. the angle ADC=BDC.

(31.) A given straight line being divided into any three parts; to determine a point such, that lines drawn to the points of section and to the extremities of the line shall contain three equal angles.

Let AB be the given line, and AC, CD, DB the

B

given parts. Take CO a mean proportional between 40 and DO; and with the centre O and radius OC describe a circle. Produce CB; and make DE a mean proportional between CE and BE; and with the centre E, and radius ED, describe a circle cutting the former in F; F is the point required. For, as was proved in the last proposition, AF FD: AC: CD,

and.. the angle AFC=CFD; and

CF: FB: CD: DB,

... the angle CFD=DFB;

and.. the three angles AFC, CFD, DFB are equal.

(32.) If two equal lines touch two unequal circles, and from the extremities of them lines containing equal angles be drawn cutting the circles, and the points of section joined; the triangles so formed will be reciprocally proportional.

Let two equal lines AB, CD touch two unequal

B

L

D

M

443

circles EBF, GDH; and from A and C let lines AIK, AEF, CLM, CGH be drawn containing the equal angles KAF, MCH. Join IE, KF, GL, MH; then will the triangle AKF: CHM :: CGL : AIE.

Since AB is equal to CD, the rectangles EA, AF, and GC, CH are equal;

.. AF: CH :: GC : AE,

and for the same reason,

whence AF × AK: CH× CM :: CG × CL : AE × AI, .. the triangle AKF : CMH :: CGL : AIE, since AK CM in the ratio of the perpendiculars from K and M on AF and CH; and CL; AI in the ratio of the perpendiculars from L and I.

(33.) If from an angle of a triangle a line be drawn to cut the opposite side, so that the rectangle contained by the sides including the angle, be equal to the rectangle contained by the segments of the side together with the square of the line so drawn; that line bisects the angle.

From B one of the angles of the triangle ABC, let BD be drawn, so that the rectangle AB, BC may be equal to the rectangle AD, DC together with the square of BD; BD bisects the angle B.

B

E

D

G F

For if not, let BE bisect it; the rectangle AB, BC is equal to the rectangle AE, EC together with the square of BE. About ABC describe a circle, and produce BD, BE to the circumference in F and G; join FG. The rectangle AD, DC is equal to the rectangle BD, DF; .. the rectangle AB, BC is equal to the rectangle BD, DF together with the square of BD, i. e. to the rectangle BF, BD. In the same way the rectangle AB, BC is equal to the rectangle BG, BE;

whence the rectangle BG, BE is equal to the rectangle BF, BD; a circle may therefore be described through D, E, G, F; whence DEG, DFG are equal to two right angles, i. e. to DEG, DEB; and .. DFG is equal to DEB, or to DAB and ABG; and .. the arc AB equal to the arc BC, which is absurd, unless the triangle be isosceles. Hence .. BG does not bisect the angle; and no other but BD can bisect it.

(34.) In any triangle, if perpendiculars be drawn from the angles to the opposite sides; they will all meet in a point.

Let ABC be any triangle; and AF, CD perpendiculars drawn upon the opposite sides, intersecting each other in G. Through G draw BGE; it is perpendicular to AC.

Join FD; and about the trapezium

BFGD describe a circle. The triangles ADG, GFC being equiangular,

AG GC GD: FG,

whence also the triangles AGC, FGD are equiangular; and .. the angle_ACD=DFG=ABE; and the angle BAC is common to the two triangles ABE, ACD; .. the angle AEB = ADC, i. e. it is a right angle, and BE is perpendicular to AC.

(35.) If from the extremities of the base of any triangle, two perpendiculars be let fall on the line bisecting the vertical angle; and through the points where they

meet that line, and the point in the base whereon the perpendicular from the vertical angle falls, a circle be described; that circle will bisect the base of the triangle.

H

Let ABC be a triangle, whose vertical angle B is bisected by the line BD, on which let fall the perpendiculars AF, CE. From B let fall BG perpendicular to AC; a circle described passing through E, F, G will also bisect AC.

D

About the triangle ABC describe the circle ADB; and from D draw a diameter which will bisect AC in H. Now since the angle AID is common to the triangles AIF, HID, and the angles AFI, IHD are right angles, .. the triangles AIF, HID are similar. same manner BIG, CEI are similar. Whence HI: ID :: IF : IA,

In the

:. HI× IG : ID × IB :: IF × IE: 1A× IC, and since ID × IB=IA × IC, .. HI× IG=IF × IE, or a circle passing through E, F, G will pass through H (vi. 13.), and .. bisect the base AC.

(36.) If from one of the angles of a triangle a straight line be drawn through the centre of its inscribed circle, and a perpendicular be drawn to this line from one of the other angles; the point of intersection of the perpendicular, and the two points of contact of the inscribed circle, which are adjacent to the remaining angle, are in the same straight line.

Let ABC be a triangle, and O the centre of its inscribed circle. From B draw BOD through the centre;