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.
Statement:
If 
then 
is on the perpendicular bisector of 
.
Proof:
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").
Converse:
If 
is the perpendicular bisector of 
then 
.
Proof:
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.
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Date last modified: August 4, 2008.