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# $ABCD$ is a square, $X$ and $Y$ are points on sides $AD$ and $BC$ respectively such that $AY = BX$. Prove that $BY = AX$ and $\angle BAY = \angle ABX$.

Given:

$ABCD$ is a square, $X$ and $Y$ are points on sides $AD$ and $BC$ respectively such that $AY = BX$.

To do:

We have to prove that $BY = AX$ and $\angle BAY = \angle ABX$.

Solution:

In $\triangle BAX$ and $\triangle ABY$,

$AB =AB$ (Common)

$BX = AY$

Therefore, by RHS axiom,

$\triangle BAX \cong \triangle ABY$

This implies,

$AX = BY$ (CPCT)

$\angle ABX = \angle BAY$ (CPCT)

**Hence, $BY = AX$ and $\angle BAY = \angle ABX$.**

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