In $\Delta PQR$,$X$ is the midpoint of $QR$. $XY$ and $XZ$ are altitudes from $X$ on $PQ$ and $PR$ respectively. If $XY = XZ$,then prove that $\Delta PQR$ is an isosceles triangle.

(N/A) Given: In $\Delta PQR$,$X$ is the midpoint of $QR$,so $QX = XR$. $XY \perp PQ$ and $XZ \perp PR$. Also,$XY = XZ$.
Step $1$: Consider $\Delta QXY$ and $\Delta RXZ$.
Step $2$: In these two triangles:
$1$. $\angle XYQ = \angle XZR = 90^{\circ}$ (Given as altitudes).
$2$. $XY = XZ$ (Given).
$3$. $QX = XR$ (Given as $X$ is the midpoint of $QR$).
Step $3$: By $RHS$ congruence criterion,$\Delta QXY \cong \Delta RXZ$.
Step $4$: By $CPCT$,$\angle Q = \angle R$.
Step $5$: In $\Delta PQR$,since $\angle Q = \angle R$,the sides opposite to these angles must be equal,i.e.,$PR = PQ$.
Conclusion: Since two sides of $\Delta PQR$ are equal,$\Delta PQR$ is an isosceles triangle.