$A B C D$ is a trapezium with $A B ∥ D C$ . A line parallel to $A C$ intersects $A B$ at $X$ and $B C$ at $Y$ . Prove that $a r (△ A D X) = a r (△ A C Y)$

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Solution

Given : $A B C D$ is a trapezium with $A B ∥ D C$ and $X Y ∥ A C .$

Join $Y A, X C$ and $X D .$

Triangles $A C X$ and $A C Y$ have same base $A C$ and are between same parallels $A C$ and $X Y$

So, $a r (△ A C X) = a r (△ A C Y)$ ......(i)
Triangles $A C X$ and $A D X$ have same base $A X$ and are between same parallels $A B$ and $D C .$
So, $a r (△ A C X) = a r (△ A D X)$ ......(ii)

From (i) and (ii),
$a r (△ A D X) = a r (△ A C Y)$

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