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.