Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

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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