Diagonals AC and BD of a quadrilateral ABCD intersect each other at P Show that ar (APB) x ar (CPD) - Maths - Quadrilaterals

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

Ankush Jain · Accepted Answer

Given, ABCD is a quadrilateral in which diagonals AC and BD of a quadrilateral ABCD intersect each other at P To prove: ar (APB) Ã— ar (CPD) = ar (APD) Ã— ar (BPC) Construction: Draw AM âŠ¥ BD and CN âŠ¥ BD

Proof: LHS = ar (APB) Ã— ar (CPD) $= \frac{1}{2} (B P \times A M) \times \frac{1}{2} (D P \times C N) = \frac{1}{2} (B P \times C N) \times \frac{1}{2} (D P \times A M)$ = ar (BPC) Ã— ar (APD) = ar (APD) Ã— ar (BPC) = RHS [Hence proved]

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