In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that `ar (ABCDE) = ar (DeltaAPQ)`.

Explanation

Given ABCDE is a pentagon.
`BP || AC` and `EQ ||AD`.
To prove `" " ar(ABCDE) = ar (APQ)`
Proof We know that, triangles on the same base and between the same parallels are equal in area.
Here, `DeltaADQ` and `DeltaADE` lie on the same base AD and between the same parallels AD and EQ.
So, `" " ar (DeltaADQ) = ar (DeltaADE)" "` ...(i)
Similarly, `DeltaACP` and `DeltaACB` lie on the same base AC and between the same parallels AC and BP.
So, `" " ar (DeltaACP) = ar (DeltaACB)" "` ...(ii)
On adding Eqs. (i) and (ii), we get
`ar (DeltaADQ) + ar (DeltaACP) = ar (DeltaADE) + ar (DeltaACB)`
On adding `ar (DeltaACD)` both sides, we get
`ar (DeltaADQ) + ar (DeltaACP) + ar (DeltaACD) = ar (DeltaADE) + ar (DeltaACB) + ar (DeltaACD)`
`rArr" "` `ar (DeltaAPQ) = ar (ABCDE)" "` Hence proved.