# Proof 3.42

- Author:
- Kayla Moore

Prove that the perpendicular bisectors of the sides of a triangle are concurrent.

Proof: Consider and let be the point of concurrency of the perpendicular bisectors of and by construction.
Consider . Since the is the perpendicular bisector of then is the midpoint of , we know that and are equal by Proposition 10. From the construction, we also know that and are equal to 90 degrees. We also know that is that same in both triangles because Common Notion 1 says that equals are equal. This being said, by Proposition 4, we know that and are congruent to one another.
Now consider . Let be the bisecting point of by construction. Since is the perpendicular bisector by Proposition 10, we know that and are equal in length and that and are 90 degrees. We also know that is equal by Common Notion 1. By Proposition 4, we know that .
Finally, consider . Let be the bisection point of . For similar reasons as above, and are equivalent in length. is equal to itself by Common Notion 1 which tells us that equals are equal. by Proposition 4.
Proposition 8 tells us that the angles enclosed in the congruent triangles are also congruent. Therefore, . Since Proposition 13 tells us that a line standing on a line creates two angles whose sum is equal to two right angles, we know . So, and because the two angles are congruent. Since , it is the perpendicular bisector.
Therefore, is concurrent to all of the triangles in question and the perpendicular bisectors of each side of the .