In a trapezium $A B C D,$ $A B ∥ D C$ and $L$ in the mid-point of $B C .$ Through $L,$ a line $L M$ is drawn parallel to $A D$ which meets $A B$ in $X$ and $D C$ produced in $Y .$ prove that $a r (A B C D) = a r (A X Y D)$

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Solution

In $△ B X L$ and $△ C L Y,$
$∠ Y C L = ∠ X B L$ [Alternate angles]
$B L = C L$ [ $L$ is the mid point of $B C$ ]
$∠ C L Y = ∠ B L X$ [Vertically Opposite angle]

So, by $A S A$ rule of congruence,
$△ B X L ≅ △ C L Y$
$\Rightarrow a r (△ C L Y) = a r (△ B L X) \dots (i)$

$a r (A B C D) = a r (D C L X A) + a r (△ X B L) \dots (i i)$

Also,
$a r (A X Y D) = a r (D C L X A) + a r (△ C Y L)$
$a r (A X Y D) = a r (D C L X A) + a r (△ B X L)$ [From $(i)$ ]
$a r (A X Y D) = a r (A B C D)$

Hence, proved.

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