Q. 245.0( 1 Vote )

# In Fig. 10.86, if quadrilateral PQRS circumscribes a circle, then PD + QB =A. PQB. QRC. PRD. PS

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

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