Two circles intersect each other at A and B as shown in the figure. The common chord AB is produced to meet a common tangent PQ to the circles at D. Prove that DP=DQ.
It follows from the alternate segment theorem that the triangles PDB and ADP are similar. Deduce that $\displaystyle PD^2 = AD*BD$. (In fact, this is a well-known theorem: from a point D, draw a tangent DP to a circle, and a chord DBA cutting the circle; then the square of PD is equal to the product of DA and DB.)
Now do the same for the other circle, to see that $\displaystyle QD^2$ is also equal to $\displaystyle AD*BD$. Conclusion: PD = QD.