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

(N/A) Given that $BM \perp AC$ and $CN \perp AB$.
Therefore,$\angle BMC = 90^{\circ}$ and $\angle BNC = 90^{\circ}$.
Since $\angle BMC = \angle BNC = 90^{\circ}$,the points $M$ and $N$ subtend equal angles at the line segment $BC$ on the same side.
According to the theorem,if a line segment joining two points subtends equal angles at two other points on the same side of the line,then the four points are concyclic.
Hence,the points $B, C, M$ and $N$ are concyclic.