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.

Start with this diagram from the equilateral triangle construction:

Constructed equilateral triangle

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.

constructed perpendicular bisector

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.

completed perpendicular bisector

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.

A point F on the perpendicular bisector of AB

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.

A point G equidistant from A and B

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

Constructing the bisector of AB through G.

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.


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Posted in Euclidean geometry, geometry
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