Parallelograms on the Same Base and Between Same Parallels
In the adjoin...
Question
In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of △ABC. If PQ∥BC and CDP and BEQ are straight lines, then prove that ar(△ABQ)=ar(△ACP).
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Solution
In △PAC PA∥DE and E is the midpoint of AC So, D is the midpoint of PC
By Converse of midpoint theorem Also, DE=12PA ----- (1) Similarly, DE=12AQ -------(2) From (1) and (2) we have PA=AQ △ABQ & △ACP are on same base PQ and between same parallels PQ and BC ∴ar(△ABQ)=ar(△ACP) (proved)