Two circles touching internally at O. OXY, OAB straight lines, the latter passing through the centres. Prove that OX : OY = OA : OB.

Given : Two circles touching internally

at ‘O' . The line OXY touches the circles
at X and Y . The line OAB passes through
the centres of the circles.
R. T. P.: OX : OY = OA : OB
Construction: Join AX and BY.
Proof : OA is the diametre of the inner
circle.
∠OXA = 90⁰ ( Angle in semicircle )......................( i )
OB is the diameter of the outer circle.

∠OYB = 90⁰ ( Angle in the semi circle...................( ii )

From ( i ) and ( ii ); ∠OXA = ∠OYB =90⁰

i.e., The corresponding angles are equal.
 AX // BY
In ΔOBY; AX //BY
OX/OY = OA/OB

i.e., OX : OY = OA : OB