In $Fig.$,$AB$ and $CD$ are common tangents to two circles of unequal radii. Prove that $AB = CD$.

(N/A) Given: $AB$ and $CD$ are common tangents to two circles of unequal radii.
To prove: $AB = CD$.
Construction: Produce $AB$ and $CD$ to intersect at point $P$.
Proof:
Since $PA$ and $PC$ are tangents drawn from an external point $P$ to the larger circle,their lengths are equal:
$PA = PC$ ... $(1)$
Since $PB$ and $PD$ are tangents drawn from the same external point $P$ to the smaller circle,their lengths are equal:
$PB = PD$ ... $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$PA - PB = PC - PD$
$AB = CD$
Hence proved.