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)

  • 105
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