In the give figure, X and Y are the midpoints of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).

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Solution

Given: Y and X are midpoints of AB and AC, respectively. QP || BC and CYQ and BXP are straight lines.
To prove: ar(∆ABP) = ar(∆ACQ)
Proof:
X and Y are the mid points of AC and AB, respectively.
So, XY || BC
In ∆BYC and ∆AYQ, we have:

$B Y = A Y (Y is the midpoint of A B) ∠ Q A Y = ∠ C B Y (Alternate angles) ∠ A Y Q = ∠ B Y C (Vertically opposite angles) S o, △ B Y C ≅ △ A Y Q (By ASA congruency)$
∴ BC = AQ ...(ii) (CPCT)

In ∆BXC and ∆AXQ, we have:

$C X = X A (X is the mid point of A C) ∠ P A X = ∠ B C X (Alternate angles) ∠ A X P = ∠ C X A (Vertically opposite angles) S o, △ B X C ≅ △ A X Q (By ASA congruency)$
∴ BC = AP ...(ii) (CPCT)

From (i) and (ii), we get:
AQ = AP
Now, ∆ABP and ∆ACQ are on the equal base (AQ = AP) and between the same parallels.
∴ ar( ∆ABP) = ar(∆ACQ)

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