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

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Solution

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

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