# Perpendicular Bisectors

After you construct an equilateral triangle, it’s natural to construct a perpendicular bisector. A perpendicular bisector of a segment is a line that intersects the segment at a right angle and cuts it into two equal parts.

Create point $D$ at the remaining circle intersection. Drawing in $AD$ and $BD$ creates another equilateral triangle congruent to the first. Construct line $CD$. This line is the perpendicular bisector of segment $AB$, as we’ll show.

We need to demonstrate that segment $AE \cong BE$ and that $\angle BEC$ is a right angle.

We have congruent triangles $ACD \cong BCD$ by SSS, meaning $\angle ACE \cong \angle BCE$.

That means triangles $ACE \cong BCE$ by SAS, and therefore $AE \cong BE$. Further, $\angle AEC \cong \angle BEC$, and since these angles are supplementary, they are both right. That completes the proof.

Next we would like to prove that a point lies on the perpendicular bisector if and only if it is equidistant from $A$ and $B$.

First we’ll take a point on the bisector and show it’s equidistant. Choose some point $F$ and draw $AF$ and $BF$.

By SAS, we have congruent triangles $AEF \cong BEF$. Therefore $AF \cong BF$.

Now we’ll choose a point equidistant from $A$ and $B$ and show it’s on the bisector. Erase the bisector and choose a point $G$ so that $AG \cong BG$.

Draw in segment $EG$. This is a bisector. It remains to show it’s perpendicular.

By SSS, $AGE \cong BGE$ so $\angle AEG \cong \angle BEG$ and since they’re supplementary they’re right, as before. The bisector $EG$ is perpendicular, so the point $G$ lies on the perpendicular bisector.

BOOM!

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1. […] I drew two perpendicular bisectors, […]