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  • In Fig., 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)

In Fig., 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)

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Solution.

Given: a pentagon ABCDE, BP \parallel AC and EQ\parallel AD.
To prove:ar(ABCDE) = ar(\triangle APQ)
Proof: We know that triangles on the same base and between the same parallel lines are equal in area.
Now, \triangle ADQ and \triangle ADE lie on the same base AD and lie between the same parallels AD and EQ. So,
ar(\triangle ADE) = ar(\triangle ADQ)…(i)
Similarly,
\triangle CAB and \triangle CAP lie on the same base AC and lie between the same parallels AC and PB
ar(\triangle CAB)=ar(\triangle CAP) …(ii)
On adding equations (i) and (ii), we get
ar (\triangle ADE) + ar(\triangle CAB) = ar(\triangle ADQ) + ar (\triangle CAP)
Adding ar (DCDA) to both sides, we get

ar(\triangle ADE)+ar(\triangle CAB)+ar(\triangle CDA)=ar(\triangle ADQ)+ar(\triangle CAP)+ar(\triangle CDA)
\Rightarrowar(ABCDE) = ar(\triangle APQ)
Hence proved.

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