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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).

Q: 6    Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show   that  ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).

            [Hint: From A and C, draw perpendiculars to BD.]

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Given that,
A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw AM \perp BD and CN \perp BD

Now, ar(\DeltaAPB) = \frac{1}{2}\times BP\times AM and,
ar(\Delta CDP)=\frac{1}{2}\times DP\times CN
Therefore, ar(\DeltaAPB) \times ar(\DeltaCDP)=
                                                         \\=(\frac{1}{2}\times BP\times AM).(\frac{1}{2}\times DP \times CN)\\ =\frac{1}{4}\times BP \times DP\times AM\times CN....................(i)

Similarly, ar(\Delta APD) \times ar (\Delta BPC) =
                                                           =\frac{1}{4}\times BP \times DP\times AM\times CN................(ii)

From eq (i) and eq (ii), we get

ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).
Hence proved

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