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

(N/A) Given: Common tangents $AB$ and $CD$ to two circles intersect at $E$.
To prove: $AB = CD$.
Proof:
From point $E$,$EA$ and $EC$ are tangents to the left circle. Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$EA = EC$ ......$(i)$
Similarly,from point $E$,$EB$ and $ED$ are tangents to the right circle. Therefore:
$EB = ED$ .......$(ii)$
Adding equations $(i)$ and $(ii)$,we get:
$EA + EB = EC + ED$
Since $EA + EB = AB$ and $EC + ED = CD$,we have:
$AB = CD$
Hence proved.