Prove that through a given point,we can draw only one perpendicular to a given line.

(N/A) Let $AB$ be a given line and $P$ be a point outside the line.
From point $P$,draw a perpendicular $PM$ to the line $AB$,such that $\angle PMB = 90^{\circ}$.
Suppose,if possible,we can draw another perpendicular $PN$ to the line $AB$,such that $\angle PNB = 90^{\circ}$.
In $\triangle PMN$,since $\angle PMB = 90^{\circ}$ and $\angle PNB = 90^{\circ}$,we observe that if $N$ and $M$ are distinct points,then $\triangle PMN$ would have two angles equal to $90^{\circ}$,which implies the sum of angles in the triangle would exceed $180^{\circ}$. This contradicts the angle sum property of a triangle.
Therefore,the points $M$ and $N$ must coincide.
Hence,through a given point,we can draw only one perpendicular to a given line.