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.

Start with this diagram from the equilateral triangle construction:

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

We need to demonstrate that segment and that is a right angle.

We have congruent triangles by SSS, meaning .

That means triangles by SAS, and therefore . Further, , 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 and .

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

By SAS, we have congruent triangles . Therefore .

Now we’ll choose a point equidistant from and and show it’s on the bisector. Erase the bisector and choose a point so that .

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

By SSS, so and since they’re supplementary they’re right, as before. The bisector is perpendicular, so the point lies on the perpendicular bisector.

BOOM!

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