Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show thatar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint: From A and C, draw perpendiculars to BD]

Question

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show thatar (APB) &#xD7; ar (CPD) = ar (APD) &#xD7; ar (BPC). [Hint: From A and C, draw perpendiculars to BD]

Philoid · Accepted Answer

Let us construct AM perpendicular to BD

Now,

Area (APB)*Area (CPD)

=

Area (APD)*Area (BPC)

=

Hence,

Area (APB)*Area (CPD) = Area (APD)*Area (CPB)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show thatar (APB) × ar (CPD) = ar (APD) × ar (BPC).[Hint: From A and C, draw perpendiculars to BD]

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show thatar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint: From A and C, draw perpendiculars to BD]