Question

In the figure, two circles intersect each other in points $A$ and $B$. Seg $AB$ is the chord of both circles. The point $C$ is the exterior point of both the circles on the line $AB$. From the point $C$, tangents have been drawn to the circles touching at $M$ and $N$. Prove that $CM=CN$.

Answer 1

Line $CBA$ is a secant intersecting the circle at point $B$ and $A$ and line $CM$ is a tangent at point $M$.
$\therefore$ ${CM}^2=CB\times CA$ .....(i)
Line $CBA$ is a secant intersecting the circle at point $B$ & $A$ and line $CN$ is a tangent at point $N$.
$\therefore$ ${CN}^2=CB\times CA$ ......(ii)
From equations (i) and (ii), we have
${CM}^2={CN}^2$
Taking square roots on both sides
$\therefore CM=CN$

Hence, proved