Perpendicular from Point to Straight Line in Plane is Unique
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Theorem
Let $BC$ be a straight line in the plane.
Let $AM$ be perpendicular to $BC$ at $M$.
Let $P$ be an arbitrary point on the same side of $BC$ as $A$ is, but not on $AM$.
Then $PM$ cannot be perpendicular to $BC$.
Proof
Without loss of generality let $M$ bisect $BC$.
Aiming for a contradiction, suppose $PM \perp BC$.
By definition $PM$ is a perpendicular bisector of $BC$ at $M$.
Then by definition of perpendicular bisector:
- $PC = PB$
Find $P'$ on $AM$ such that $P'C = PC$.
Since $AM$ is also a perpendicular bisector of $BC$:
- $P'B = P'C$
- $\leadsto PB = PC = P'C = P'B$
In the words of Euclid:
- Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot constructed on the same straight line (from its extremities) and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.
(The Elements: Book $\text{I}$: Proposition $7$)
This is a contradiction.
Therefore, no such $P$ exists.
All points equidistant from $B$ and $C$, on the same side of $BC$ as $A$, lie on $AM$.
$\blacksquare$