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# Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD)×ar (BPC).

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

ar (APB) × ar (CPD) = ar (APD) × ar (BPC).

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Given,

The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.

Construction:

From A, draw AM perpendicular to BD

From C, draw CN perpendicular to BD

To Prove,

ar(ΔAED) ar(ΔBEC) = ar (ΔABE) × ar (ΔCDE)

Proof:

ar(ΔABE) = 1/2 × BE × AM………….. (i)

ar(ΔAED) = 1/2 × DE × AM………….. (ii)

Dividing eq. (ii) by (i), we get,

$$\frac{ar(\triangle AED)}{ar(\triangle ABE)}$$ = $$\frac{\frac{1}{2} \times DE \times AM}{\frac{1}{2} \times BE \times AM}$$

ar(AED)/ar(ABE) = DE/BE…….. (iii)

Similarly,

ar(CDE)/ar(BEC) = DE/BE ……. (iv)

From eq. (iii) and (iv), we get

ar(AED)/ar(ABE) = ar(CDE)/ar(BEC)

ar(ΔAED) × ar(ΔBEC) = ar(ΔABE) × ar (ΔCDE)

Hence proved.

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