Question

In the given figure, common tangents $AB$ and $CD$ to the two circles with centres $O_1$ and $O_2$ intersect at $E.$ Prove that $AB=CD.$

Answer 1

We know that tangent segments to a circle from the same external point are congruent.

So, we have

$EA=EC$ for the circle having centre $O_1$

And, $ED=EB$ for the circle having centre $O_2$

Now, Adding $ED$ on both sides in equation $EA=EC,$ we get

$EA+ED=EC+ED$

$\Rightarrow EA+EB=EC+ED$ [$\because ED=EB$]

$\Rightarrow AB=CD$

Or

Common tangents $AB$ and $CD$ to the two circles with centres $O_1$ and $O_2$ intersect at $E.$

$EA=EC----\left(1\right)$ [Tangents drawn from an external point to a circle are equal]

$EB=ED-----\left(2\right)$ [Tangents drawn from an external point to a circle are equal]

Adding equations (1) and (2), we have

$EA+EB=EC+ED$

$AB=CD$

Hence proved