The Perpendicular Bisector Theorem
We have discussed this before, and now we will give a precise proof:
If a point is equidistant from the endpoints of a segment then it is on the perpendicular bisector of that segment, and conversely.
There are two things to prove here, a statement and its converse, so we will split the proof into two parts, using generic diagrams for each.
If then is on the perpendicular bisector of .
Let be the midpoint of and draw :
Then by SSS. Therefore since these are corresponding parts of the congruent triangles. But these angles are a linear pair, so their measures add to 180o. Since they are congruent angles, each must measure half of 180o, which is 90o. Therefore . So is on the perpendicular bisector of (by definition of "perpendicular bisector").
If is the perpendicular bisector of then .
Because is the perpendicular bisector of , , and since both are right angles. is a common side to and , so these triangles are congruent by SAS. Therefore since they are corresponding parts of the congruent triangles.