Question

As shown in the given figure, two circles intersect each other in two points A and B. Seg AB is the chord of both circles. Point C is the exterior point of both the circles on the line AB.From the point C tangents are drawn to the circles touching at M and N. Prove that CM = CN.

Answer 1

Since AC is a secant and CM is a tangent and by secant-tangent theorem, we have:
$\mathrm{CB}\times \mathrm{CA}={\mathrm{CM}}^2$ ...(1)

Since AC is a secant and CN is a tangent and by secant-tangent theorem we have:
$\mathrm{CB}\times \mathrm{CA}={\mathrm{CN}}^2$ ...(2)
Using equations (1) and (2), we get:
$$\begin{gathered} {\mathrm{CN}}^2={\mathrm{CM}}^2 \\ \mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \mathrm{CN}=\mathrm{CM} \\ \mathrm{Hence},\mathrm{proved} \end{gathered}$$