In a figure $X$ and $Y$ are the mid point of $A C$ and $A B$ respectively, $Q P ∥ B C$ and $C Y Q$ and $B X P$ are straight lines. Prove that $a r (△ A B P) = a r (△ A C Q)$ .

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Solution

Since $X$ and $Y$ are the mid points $A C$ and $A B$ respectively.

$∴ X Y ∥ B C$

Clearly, triangles $B Y C$ and $B X C$ are on the same base $B C$ and between the same parallel $X Y$ and $B C$

$∴ a r (△ B Y C) = a r (△ B X C)$

$\Rightarrow a r (△ B Y C) - a r (△ B O C) = a r (△ B X C) - a r (△ B O C)$

$\Rightarrow a r (△ B O Y) = a r (△ C O X)$

$\Rightarrow a r (△ B O Y) + a r (△ X O Y) = a r (△ C O X) + a r (△ X O Y)$

$\Rightarrow a r (△ B X Y) + a r (△ C X Y) . . . (i)$

$\Rightarrow a r (△ B Y X) + a r (△ A X Y) = a r (△ A X Y) + a r (△ A Y X)$

$\Rightarrow a r (△ B A X) = a r (△ C A Y) . . . (i i)$

In $△ B A P, X Y = \frac{1}{2} A P$

In $△ C A Q, X Y = \frac{1}{2} A Q$

$∴ A P = A Q$

Clearly, triangles $X A P$ and $Y A Q$ are between the same parallels and their bases $A P$ and $A Q$ are equal.

$∴ a r (△ X A P) = a r (△ Y A Q) . . . . . (i i i)$

Adding $(i i)$ and $(i i i)$ , we obtain

$a r (△ X A P) + a r (△ B A X) = a r (△ Y A Q) a r (△ C A Y)$

$\Rightarrow a r (△ A B P) = a r (△ A C Q)$

Hence Proved

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