Perpendicular bisector theorem

Under the assumptions of the Common Core, the triangle congruence criteria (SSS, SAS, ASA) are no longer postulates. Instead they are to be derived from transformational geometry. There are many consequences to this state of affairs, and I have written about it on this blog: one,  two, three. I have developed bits of curriculum, which you can download here, and taught summer workshops on the topic. (Next one coming up in August in Seattle!)

A key ingredient in proving the congruence criteria is this theorem:

If a point P is equidistant from two points A and B, it lies on their perpendicular bisector.

The theorem is not hard to prove using congruent triangles, but that's not an option, since we need it to prove the congruence criteria. I included a transformational proof in my article "Triangle Congruence and Similarity A Common-Core-Compatible Approach" (Version 1, Version 2). While that proof is correct, I found that the teachers I presented it to found it confusing. So here is an alternate proof, which is perhaps more readily understandable. (It relies on the definitions and assumptions made in the article.)

Given: P is equidistant from A and B
Prove: it lies on their perpendicular bisector.

 

Draw the angle bisector b of ∠APB. If we can show that b is the perpendicular bisector of AB, then we are done, since P is on it.

If we were to reflect A in b, where would its image A' be? Since reflections preserve angles, A' must be on the ray PB. Since reflections preserve distance, A' must be on the circle centered at P, with radius PA. But the intersection of the ray and the circle is B, so A'=B. It follows that B is the reflection of A in b. Therefore b is the perpendicular bisector of AB. QED.

Is this clearer? Let me know.

--Henri