$ABCD$ 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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