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ABCD is a rhombus. P is any point on BC, AP is joined and produced to meet DC produced at Q. Prove that (1/BC)=(1/PC)-(1/CQ)

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(1) angle APB = angle angle CPQ (vertically opposite angles)

(2) Lines AB and CQ are parallel (CQ being part of side D extended)

and line BC intersects them.

Therefore alternate angles PAB and PQC are equal.

(3) Also for the same reason as at (2),

alternate angles PBA and PCQ are equal.

Hence triangles PAB and PQC are similar.

Therefore, PB/AB = PC/CQ --------(4)

As PB = BC - PC, and AB = BC, above equation may be written as

(BC-PC) / BC = PC / CQ or 1 - (PC / BC) = PC/CQ

OR (PC/BC) + (PC/CQ) = 1 or PC[(1/BC) + (1/CQ)] = 1

OR (1/BC) + (1/CQ) = 1/PC

OR 1/BC = (1/PC) - (1/CQ)