ABCD is a square; X is the mid-point of AB and Y the mid-point of BC .
Hence, The triangles ADX and BAY are congruent
If the above statement is true then mention answer as 1, else mention 0 if false

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Solution

$A B = B C$ (ABCD is a square)
$\frac{1}{2} A B = \frac{1}{2} B C$
$A X = B Y$ (X and Y are mid points of AB and BC respectively)
In $△ A X D$ and $△ A B Y$ ,
$∠ D A X = ∠ A B Y$ (Each $90^{\circ}$ )
$A X = B Y$ (Proved above)
$A D = A B$ (ABCD is a square)
Thus, $△ A X D ≅ △ B Y A$

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