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Question
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that : PA x PD = PB x PC.
Solution
In ΔAPB and ΔCPD,
∠APB = ∠CPD .......(vertically opposite angles)
∠ABP = ∠CDP ....(alternate angles since AC || DC)
ΔAPB ∼ ΔCPD .....(AA criterion for similarity)
`=> (PA)/(PC) = (PB)/(PD)` .....(Since corresponding sides of similar triangles are equal)
`=>` PA x PD = PB x PC
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