Question

$AB$ and $CD$ are common tangents to two circles of equal radii. Prove that $AB=CD$.

Answer 1

In this figure,

$OA=OD=PC=BP\left[Radius\right]$

or $OA+OD=PC+BP$

$AD=BC$ [Diameter]

$\angle BAO=\angle ODC=\angle DCP=\angle PBA={90}^o$

[Radius is always perpendicular to the tangent]

So, ABCD is a rectangle.

Hence AB$=$CD

Hence proved.