Question

The diagonals AC and BD of the quadrilateral ABCD intersect at P. Prove that ar(∆APD) × ar(∆BPC) = ar(∆APB) × ar(∆CPD)

Answer 1

area of triangle= 1/2×base×altitudine;
so:

ar(apb)×ar(cpd)=(1/2×bp×am)×(1/2×pd×cn)
                         = 1/4×bp×am×pd×cn

ar(apd)×ar(bpc)=(1/2×pd×am)×(1/2×cn×bp)
                         =1/4×pd×am×cn×bp
                         =1/4×bp×am×pd×cn

then
ar(apb)×ar(cpd)=ar(apd)×ar(bpc)