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 circlesat X and Y . The line OAB passes throughthe centres of the circles.R. T. P.: OX : OY = OA : OBConstruction: Join AX and BY.Proof : OA is the diametre of the innercircle.∠OXA = 90^{0} ( Angle in semicircle )......................( i )OB is the diameter of the outer circle.
∠OYB = 90^{0} ( Angle in the semi circle...................( ii )
From ( i ) and ( ii ); ∠OXA = ∠OYB =90^{0}
i.e., The corresponding angles are equal. AX // BYIn ΔOBY; AX //BYOX/OY = OA/OB
i.e., OX : OY = OA : OB