In the adjoining figure,D and E are respectively the midpoints of sides AB and AC of $△ A B C$ .If PQ||BC and CDP and BEQ are straight lines then prove that $a r (△ A B Q) = a r (△ A C P)$ .

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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=1/2PA-------(1)

Similarly, DE=1/2AQ-------(2)
From (1) and (2) we have
PA = AQ
∆ABQ and ∆ACP are on same base PQ and between same parallels PQ and BC
ar(∆ABQ) = ar(∆ACP)

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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).

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=1/2PA-------(1) Similarly, DE=1/2AQ-------(2) From (1) and (2) we have PA = AQ ∆ABQ and ∆ACP are on same base PQ and between same parallels PQ and BC ar(∆ABQ) = ar(∆ACP)

In the adjoining figure,D and E are respectively the midpoints of sides AB and AC of $△ A B C$ .If PQ||BC and CDP and BEQ are straight lines then prove that $a r (△ A B Q) = a r (△ A C P)$ .

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=1/2PA-------(1)

Similarly, DE=1/2AQ-------(2)
From (1) and (2) we have
PA = AQ
∆ABQ and ∆ACP are on same base PQ and between same parallels PQ and BC
ar(∆ABQ) = ar(∆ACP)