If $BM$ and $CN$ are the altitudes drawn on the sides $AC$ and $AB$ of the triangle $ABC$,prove that the points $B, C, M$ and $N$ are concyclic.

(N/A) $1$. In $\triangle ABC$,$BM \perp AC$ and $CN \perp AB$.
$2$. Therefore,$\angle BMC = 90^\circ$ and $\angle BNC = 90^\circ$.
$3$. Consider the line segment $BC$ as a chord.
$4$. The angles subtended by the line segment $BC$ at points $M$ and $N$ are $\angle BMC = 90^\circ$ and $\angle BNC = 90^\circ$.
$5$. Since $\angle BMC = \angle BNC = 90^\circ$,the points $M$ and $N$ lie on a circle with $BC$ as the diameter.
$6$. Thus,the points $B, C, M$ and $N$ are concyclic.