Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
Given,
The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.
Construction:
From A, draw AM perpendicular to BD
From C, draw CN perpendicular to BD
To Prove,
ar(ΔAED) ar(ΔBEC) = ar (ΔABE) × ar (ΔCDE)
Proof:
ar(ΔABE) = 1/2 × BE × AM………….. (i)
ar(ΔAED) = 1/2 × DE × AM………….. (ii)
Dividing eq. (ii) by (i), we get,
\(\frac{ar(\triangle AED)}{ar(\triangle ABE)}\) = \(\frac{\frac{1}{2} \times DE \times AM}{\frac{1}{2} \times BE \times AM}\)
ar(AED)/ar(ABE) = DE/BE…….. (iii)
Similarly,
ar(CDE)/ar(BEC) = DE/BE ……. (iv)
From eq. (iii) and (iv), we get
ar(AED)/ar(ABE) = ar(CDE)/ar(BEC)
ar(ΔAED) × ar(ΔBEC) = ar(ΔABE) × ar (ΔCDE)
Hence proved.