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About LinkBacksTwo 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 thatOriginally Posted by snigdha
. (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 thatis also equal to
. Conclusion: PD = QD.
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