In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ∆ABC such that AX = AY.

In $\triangle \mathrm{AXC}$ and $\triangle \mathrm{AYB}$, we have :

$\mathrm{AC}=\mathrm{AB}$ (Given)

$\mathrm{AX}=\mathrm{AY}$ (Given)

$\angle \mathrm{BAC}=\angle \mathrm{CAB}$ (Angle common to $\triangle \mathrm{AXC}$ and $\triangle \mathrm{AYB}$ )

$\therefore \triangle \mathrm{AXC} \cong \triangle \mathrm{AYB} \quad$ (SAS criterion)

So, $\mathrm{CX}=\mathrm{BY} \quad(\mathrm{CPCT})$

Hence, proved.