In Fig. 10.86, if quadrilateral PQRS circumscribes a circle, then PD + QB =
Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
By the above property,
PD = PA (tangent from P)
QB = QA (tangent from Q)
RC = RB (tangent from R)
SC = SD (tangent from S)
Now,
PD + QB = PA + QA
⇒ PD + QB = PQ [∵PQ = PA + QA]
Hence, PD + QB = PQ