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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 :
- ΔAPB is similar to ΔCPD.
- PA × PD = PB × PC.
Solution
In trapezium ABCD
AB || DC
Diagonals AC and BD intersect each other at P.
To prove:
- ΔAPB ∼ ΔCPD.
- PA × PD = PB × PC.
Proof: In ΔAPB and ΔCPD
∠APB = ∠CPD ...(Vertically opposite angles)
∠PAB = ∠PCD ...(Alternate angles)
∴ ΔAPB ∼ ΔCPD ...(AA axiom)
∴ `(PA)/(PC) = (PB)/(PD)`
`\implies` PA × PD = PB × PC
Hence proved.
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