Proof
But that's the theorem. Can we prove it? Here's a line, AB, and its perpendicular bisector. Let's label a point on the bisector C and the spot it hits AB as M.
Perpendicular bisector C intersects line AB at point M
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We're trying to prove that the lines we could draw from C to A and C to B are congruent. Adding those lines gives us more than just a place to hang our laundry, it gives us two triangles. That'll be important later.
Let's start by stating that CM is a perpendicular bisector of AB. We're given that. And, well, that's all we're given. But trust me, that's all we need.
Okay, now let's state that M is the midpoint of AB. That's part of the definition of a perpendicular bisector; it would be the bisector part of that definition. CM bisects AB, so the place where it hits AB is the exact middle. That means that AM is congruent to MB.
Why? Well, that's what a midpoint is - the spot that divides a line into two congruent segments. It also allows us to have pretty, symmetrical gardens on either side of our tower, or parking. Nothing beats ample parking.
Next, let's state that angles CMA and CMB are right angles. No leaning tower here. This is the other part of the definition of a perpendicular bisector; it's the perpendicular part, of course.
A perpendicular bisector creates two right angles
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Since those are right angles, we can state that angles CMA and CMB are congruent. All right angles are what? 90 degrees. And what's a 90 degree angle's relationship to another 90 degree angle? They're twins!
So we have congruent sides and congruent angles. All that came from one given statement. Not bad. And we're not done yet.
Let's state that CM is congruent to CM. File that under 'obvious statements.' Our tower equals itself. Officially, it's the reflexive property.
Okay, look what we have now: side-angle-side. That's not just a geometric palindrome, it's a postulate: SAS. That means that triangle CMA is congruent to triangle CMB.
And if triangle CMA is congruent to triangle CMB, and the corresponding parts of congruent triangles are congruent, which they are and we call CPCTC, then CA is congruent to CB. Two equal laundry lines! That's our proof!
Converse
So the perpendicular bisector theorem is true. What about its converse? The converse of a statement is when you switch the hypothesis and conclusion. Instead of 'if p, then q,' the converse is 'if q, then p.'
So the perpendicular bisector theorem states 'if a point is on the perpendicular bisector of a segment, then it is equidistant from the segment's endpoints.' Let's swap that around to read 'if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.'
Imagine the floors of our building floating in space, not connected to each other as a tower. Weird, okay. But go with it. To test our converse, let's start with a segment. This is segment AB. Let's add a point and call it X.
X is equidistant from A and B
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Our converse statement tells us that if the distance from X to A is the same as the distance from X to B, then X is on the perpendicular bisector of AB.
We know we can draw a line from X that's perpendicular to AB. We could do that from any point. Here's point Y.
Y is not equidistant from points A and B
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We could do the same thing from Y. But Y isn't equidistant from A and B, so it's not on the perpendicular bisector. Plus, does the line that includes Y look like it bisects AB? No. I think Y comes from Pisa.
But what about X? Let's add lines XA and XB. Let's call the point where our line from X hits AB Z. If we were doing a proof like before, what would we know?
Perpendicular bisector X intersects line AB at point Z
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We know XA is congruent to XB since the hypothesis of our statement tells us that it's equidistant from the segment's endpoints.
We also know that we have congruent angles - those right angles, XZA and XZB. And we know XZ is congruent to itself. Therefore, we could use the hypotenuse-leg (HL) theorem to state that these two triangles are congruent. The HL theorem states that if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the triangles are congruent.
That would enable us to state that AZ is congruent to ZB. That means that XZ does bisect AB, so X is on the perpendicular bisector of AB.
Lesson Summary
To summarize, the Leaning Tower of Pisa is actually really cool. I went there once. It was kind of awesome. Oh, and the perpendicular bisector theorem - the theorem states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the segment's endpoints. We proved this to be true. We also showed how the converse of the theorem is true. The converse states that if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. Now I think I need to go take in some laundry.
Learning Outcome
At the end of this lesson, you should be able to recall the perpendicular bisector theorem and test converse statements.
