Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
[Hint: From A and C, draw perpendiculars to BD]
Let us draw AM ⊥ BD and CN ⊥ BD
Area of a triangle
∴ ar (APB) × ar (CPD) = ar (APD) × ar (BPC)
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
[Hint: From A and C, draw perpendiculars to BD]
Let us draw AM ⊥ BD and CN ⊥ BD
Area of a triangle
∴ ar (APB) × ar (CPD) = ar (APD) × ar (BPC)