Circle C(O,r) touches the circle C(o,r') internally at P. PB is a chord of larger circle, which intersects the smaller circle at A. Then prove that PA : PB = r : r'.

i) Angle included between the chord and tangent of a circle is equal to the angle in the alternate segment. Also the angle subtended by the chord at the center is twice the angle in the segment of the same chord.

ii) Thus from the above, if the common tangent at P to two circles is QPT, 
then, <AOP = 2<APT
Also <BO'P = 2<BPT

iii) Thus from the above, <AOP = <BO'P

iv) In the traingle AOP, OA = OP = r; ==> <OPA = <OAP = (1/2)*(180 - <AOP)
Similalry <O'PB = <O'BP = (1/2)*(180 - <BOP)

Thus from all the 3 above the two triangles AOP & BOP are similar (AAA similarity)

v) ==> PA/PB = OP/O'P = r/r' [Proved]