Question 9
In figure X and Y are the mid - points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that $a r (Δ A B P) = a r (Δ A C Q)$ .

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Solution

Since X and Y are the midpoints of AC and AB respectively,
XY || BC
We know that triangle on the same base and between the same parallels are equal in area.
Here, $Δ B Y C$ and $Δ B X C$ lie on same base BC an between the same parallels BC and XY.
So, $a r (Δ B Y C) = a r (Δ B X C)$
On subtracting $a r (Δ B O C)$ from both sides, we get
$a r (Δ B Y C) - a r (Δ B O C) = a r (Δ B X C) - a r (Δ B O C)$
$\Rightarrow a r (Δ B O Y) = a r (Δ C O X)$
On adding $a r (Δ X O Y)$ both sides, we get
$a r (Δ B O Y) + a r (Δ X O Y) = a r (Δ C O X) + a r (Δ X O Y)$
$a r (Δ B Y X) = a r (Δ C X Y)$ ...(i)
Hence, we observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.
$∴$ ar(XYAP) = ar (YXAQ) ....(ii)
On adding eqs. (i) and (ii), we get,
$a r (Δ B Y X) + a r (X Y A P) = a r (Δ C X Y) + a r (Y X A Q)$
$\Rightarrow a r (Δ A B P) = a r (Δ A C Q)$

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Question 9 In figure X and Y are the mid - points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(ΔABP)=ar(ΔACQ).

Question 9
In figure X and Y are the mid - points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that $a r (Δ A B P) = a r (Δ A C Q)$ .