In the adjoining 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 (∆APQ).

Given ABCDE is a pentagon, in which BP || AC and EQ || AD.

To prove ar (ABCDE) = ar (AAPQ)

Proof We know that triangles on the same base and between the same parallels are equal in area.

∴ ar (∆ADE) = ar (∆ADQ) ,…(i)

[∵∆ADQ and ∆ADE lie on the same base AD and between the same parallels AD and EQ]

Similarly, ar (∆CAB) = ar ∆(CAP) …(ii)

[∵same base is AC and same parallels are AC and BP]

On adding Eqs. (i) and (ii), we get

ar (∆ADE) + ar (∆CAB)

= ar (∆ADQ) + ar (∆CAP) (1)

On adding ar(∆CDA) both sides, we get

ar (∆ADE) + ar (∆CAB) + ar(∆CDA)

= ar (∆ADQ) + ar (∆CAP) + ar(∆CDA)

=> ar (ABCDE) = ar (∆APQ)