Proof 3.42

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

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