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Standard IX

Mathematics

SAS Criteria for Congruency

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, $DE = \frac{1}{2} PA$ .....(1)
Similarly, $DE = \frac{1}{2} AQ$ .....(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=12PA .....(1) Similarly, DE=12AQ .....(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)