How do you construct perpendicular bisectors of a triangle?

To construct perpendicular bisectors of a triangle `Delta ABC` you have to consider each side separately as a segment (`AB`, `BC` and `AC`) and construct a perpendicular bisector to each of them.

The easy way to construct a perpendicular bisector `PQ` to segment `AB` is pictured below.

Here the centers of these circles are the endpoints of a given segment `AB` and their radiuses must be the same. The only condition is for these circles is the existence two points of intersection, `P` and `Q`. For this the radius can be any, as long as it's greater than half of the length of `AB`. The simple method is to choose it to be equal to the length of `AB`.

What's more interesting is to prove that this construction delivers the perpendicular bisector.
Here is the proof.

Assume that `M` is an intersection of `AB` and `PQ`.
`AP=BP=AQ=BQ` - each is a radius, which we have chosen
`Delta APQ = Delta BPQ` - by side-side-side theorem

Hence:
`=>/_APQ=/_BPQ` as angles of congruent triangles lying across congruent sides `AQ` and `BQ`
`=> Delta APM = Delta BPM` by side-angle-side theorem
`=> AM=BM` as sides of congruent triangles lying across congruent angles `/_APM=/_BPM`
`=>/_AMP=/_BMP` as angles of congruent triangles lying across congruent sides `AP` and `BP`
`=>/_AMP=/_BMP=90^o` since their sum is `180^o`

So, we have proven that `M` is a modpoint of `AB` and `PM_|_AB`.