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 $\angle BMC = \angle BNC = 90^{\circ}$,both angles are equal.
These two equal angles are subtended by the same line segment $BC$ at points $M$ and $N$ on the same side of $BC$.
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 containing the segment,then the four points are concyclic.
Thus,the points $B, C, M$ and $N$ are concyclic.
Therefore,$\angle BMC = 90^{\circ}$ and $\angle BNC = 90^{\circ}$.
Since $\angle BMC = \angle BNC = 90^{\circ}$,both angles are equal.
These two equal angles are subtended by the same line segment $BC$ at points $M$ and $N$ on the same side of $BC$.
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 containing the segment,then the four points are concyclic.
Thus,the points $B, C, M$ and $N$ are concyclic.
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