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Question: A circle is inscribed in the quadrilateral PQRS. Prove that: QR + SP = QP + SR.
Solution:
We know that tangents from an external [same] point to a circle are equal to one another. Let this be Theorem 1.
Q is a common point for QC and QD.
∴ QC = QD → Eq(1)
R is a common point for RC and RB.
∴ CR = RB → Eq(2)
S is a common point for SA and SB.
∴ SA = SB → Eq(3)
P is a common point for PD and PA.
∴ PA = PD → Eq(4)
Adding equation 1, 2, 3 and 4 we get,
QC + CR + SA + PA = QD + RB + SB + PD
QR + SA + PA = QP + RB + SB
∴ QR + SP = QP + SR
Hence Proved!
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