Perpendicular bisector of a segment

We return to the characterisation of the perpendicular bisector as a locus, this time with a complete figure that accompanies the proof step by step.

Theorem — Perpendicular bisector of a segment

If a point $C$ is at equal distance from the endpoints $A$ and $B$ of a segment, then $C$ lies on the perpendicular bisector of $AB$.

Proof

$CA\cong CB$ and $M$ midpoint of $AB$: the triangles $ACM$ and $BCM$ are congruent by the third criterion, so $CM\perp AB$.

Success

1. Hypothesis. $AC \cong BC$ (hypothesis).
2. Construction. I mark $M$ as the midpoint of $AB$ and join $C$ to $M$.
3. Consider. The triangles $ACM$ and $BCM$:
   • $AC \cong BC$ (hypothesis)
   • $AM \cong BM$ ($M$ midpoint)
   • $CM \cong CM$ (common side)
4. Deduce. By the third criterion: $\triangle ACM \cong \triangle BCM$.
5. Deduce. It follows that $\widehat{AMC}\cong\widehat{BMC}$. Since they are supplementary and congruent, they are both right angles.
6. Thesis. Hence $CM$ is perpendicular to $AB$ and passes through the midpoint $M$: $C$ lies on the perpendicular bisector of $AB$.

$\blacksquare$

Links

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