gent at A; then will AC AB :: BC: BD.

Produce CB to meet the tangent in E. Since the angle EAB is equal to the angle

in the alternate segment ACB, and the

A

Z

angle AEB is equal to CBD, .. the triangle ABE is similar to CBD,

and AE AB :: CB: CD;

but from similar triangles BDC, EAC, AC AE: DC: DB,

:

.. ex æquo AC : AB :: CB : DB.

(27.) If a triangle be inscribed in a circle, and from its vertex lines be drawn parallel to tangents at the extremities of its base; they will cut off similar triangles.

Let ABC be a triangle inscribed in a circle, and AD, CE tangents at the points A and C. From B draw BF, BG respectively parallel to them; these lines will cut off the triangles ABF, CBG, which are similar.

A

D

B

F G

For (Eucl. iii. 32.) the angle ACB is equal to DAB, i. e. to the alternate angle ABF; and the angle BAC is equal to BCE, i. e. to CBG; whence the triangles ABF, CBG having two angles in each equal, will be equiangular and similar.

COR. 1. The rectangle contained by the segments of the base adjacent to the angles will be equal to the square of either line drawn from the vertex.

For if AD and CE be produced, they will meet and

form with AC an isosceles triangle, to which BFG is similar, .. BF=BG.

Now AF: BF :: BG: GC

the rectangle AF, GC is equal to the rectangle BF, BG, i. e. to the square of BF.

COR. 2. Those segments are also in the duplicate ratio of the adjacent sides.

For the triangles ABF and CBG are each of them similar to ABC, whence AC: AB :: AB : AF,

.. AC

AF in the duplicate ratio of AC: AB;

for the same reason,

AC CG in the duplicate ratio of AC: CB, .. AF: CG in the duplicate ratio of AB: CB.

(28.) If one circle be circumscribed and another inscribed in a given triangle, and a line be drawn from the vertical angle to the centre of the inner, and produced to the circumference of the outer circle; the whole line thus produced has to the part produced the same ratio that the sum of the sides of the triangle has to the base.

Let ABD be a circle circumscribed about the triangle ABC; O the centre of the inscribed circle. Join 40, and produce it to D; then AOD bisects the angle BAC. Join BD, DC; and draw BO, CO to the centre of the inscribed circle; then

AD: DO :: AB+ AC: CB.

B

Draw OF, OG parallel to AB, AC, meeting BD, CD in F and G. The angle DBC= DAC = DAB= DOF, and the angle at D is common to the triangles

BED, OFD, and (vii. 20.) BD = DO, .. OF=BE. In the same manner it may be shewn that OG=EC. Now the trapeziums BACD, FOGD being similar, and similarly situated,

AD: OD :: AB+AC: FO+OG

:: AB+AC: BC.

(29.) If in a right-angled triangle, a perpendicular be drawn from the right angle to the hypothenuse, and circles inscribed within the triangles on each side of it; their diameters will be to each other as the subtending sides of the right-angled triangle.

Let ABC be a right-angled triangle; from the right angle B let fall the perpendicular BD; and in the triangles ABD, BDC let circles be inscribed; their diameters are to one another as AB to BC.

Bisect the angles BAD, ABD by the lines AO, BO, they will meet in the centre 0; in the same manner lines bisecting DBC, DCB meet in the centre E; draw OF, EG to the points of contact. Now the triangles ABD, BDC being similar (Eucl. vi. 8.), ... the triangles ABO, BCE are similar; whence

AB BC: BO: CE;

but the triangles OBF, EGC are similar,

.. BOCE :: OF EG :: 2OF: 2EG,

.. AB BC: 2OF: 2 EG.

(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 0, and radius CO, describe a circle; it will be the locus required.

In the arc CD 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 E

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; Fis 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

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.