The perpendicular bisector as a locus

We prove one of the two implications that characterise the perpendicular bisector as a locus: every point equidistant from the endpoints lies on the perpendicular bisector.

Proof — The perpendicular bisector as a locus

We prove that if a point $C$ is equidistant from the endpoints $A$ and $B$ of a segment, then $C$ lies on the perpendicular bisector of $AB$.

1. Hypothesis. $CA\cong CB$.
2. Construction. Let $M$ be the midpoint of $AB$. Join $C$ to $M$.
3. Observe. The triangles $CMA$ and $CMB$ have: $CA\cong CB$ (hypothesis), $AM\cong MB$ (construction), $CM$ in common.
4. Deduce. By the third criterion (SSS): $\triangle CMA\cong\triangle CMB$.
5. Deduce. Hence $\widehat{CMA}\cong\widehat{CMB}$. But they are also supplementary ($\widehat{CMA}+\widehat{CMB}=\pi$), so $\widehat{CMA}=\widehat{CMB}=\dfrac{\pi}{2}$.
6. Deduce. It follows that $CM\perp AB$ at the midpoint $M$: that is, $C$ lies on the perpendicular bisector of $AB$.

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Perpendicular bisector · Congruence criteria · Proof · Locus
Skills: Proving · Synthetic geometry