Question

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Answer 1

Answer:

that's it, hope you have understood.

Answer 2

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!