Question:
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar(ΔBPC).
Solution:
Construction: - Draw BQ ⊥ AC and DR ⊥ AC
Proof:-
L.H.S
= ar(ΔAPB) × ar(ΔCDP)
= (1/2) [(AP × BQ)] × (1/2 × PC × DR)
= (1/2 × PC × BQ) × (1/2 × AP × DR)
= ar(ΔAPD) × ar(ΔBPC).
= R.H.S
Hence proved.
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