Question

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

Answer 1

Let us draw AM ⊥ BD and CN ⊥ BD

Area of a triangle = 1/2 x Base x Altitude

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