Question

Two circles intersect each other at $A$ and $B$. The common chord $AB$ is produced to meet common tangent $PQ$ to the circle at $D$. Prove that $\mathrm{PQ}$ is bisected at $D$.

Answer 1

From Figure $,PD$ is the tangent and $DAB$is the secant for the first circle.
${\mathrm{PD}}^2=\mathrm{DA}\times \mathrm{DB}\dots \text{(i)}$ [by tangent secant property]
Again from Figure $,DQ$ is the tangent and $DAB$ is the secant for the second circle.
$D{Q}^2=DB\times DA......\left(ii\right)$
From (i) and (ii) we get,
$P{D}^2=D{Q}^2$
$\Rightarrow PD=DQ$

Thus, $D$ is the midpoint of $PQ$.