In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC pro – InterviewSolution

InterviewSolution

1.	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)`.
Answer» 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.

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