Question
$ABCD$ is a square, $X$ and $Y$ are points on sides $AD$ and $BC$ respectively such that $\text{AY}=\text{BX}.$ prove that $\text{BY}=\text{AX}$ and $\angle\text{BAY}=\angle\text{ABX}.$
✓
Answer
In $\triangle\text{ABX}$ and $\triangle\text{BAY}$
$\text{AY}=\text{BX}$ [given] $\text{AB}=\text{AB}$ [common]
$\angle\text{BAX}=\angle\text{ABY}=90^\circ$ [given]
By $RHS$ congurence criterion $\triangle\text{ABX}\cong\triangle\text{BAY}$
$\therefore\text{AY}=\text{BX}$ [c.p.c.t]
$\angle\text{BAY}=\angle\text{ABX}$ [c.p.c.t]
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