Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

Question

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(&#x394; APB) &#xD7; ar(&#x394; CPD) = ar(&#x394; APD) &#xD7; ar(&#x394; BPC).

Philoid · Accepted Answer

Construction: Draw BQ perpendicular to AC

And,

DR perpendicular to AC

Proof: We have,

L.H.S = Area (ΔAPB) * Area (ΔCPD)

= (AP * BQ) * (PC * DR)

= ( * PC * BQ) * ( * AP * DR)

= Area (ΔBPC) * Area (ΔAPD)

= R.H.S

Therefore,

L.H.S = R.H.S

Hence, proved

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).