*Jeff Calareso*Show bio

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

Lesson Transcript

Instructor:
*Jeff Calareso*
Show bio

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

The perpendicular bisector theorem asserts that the point of perpendicular bisection of a segment is equidistant from either end. Examine the proofs, and the converse of this theorem through the provided examples.
Updated: 09/30/2021

Let's talk perpendicular bisectors. These lines are immensely useful. Let's say you're an architect. I wanted to be an architect once. Then I realized it was less building cool models and more building regulations and codes. Anyway, let's completely ignore building codes here.

Here's some land. You're designing a skyscraper. The folks who hired you made two demands: it needs to go straight up and it needs to be in the middle of the land. Seems simple enough, right?

Say, do you know what you did? You made a perpendicular bisector. And it's not as simple as it seems. Consider the Leaning Tower of Pisa. Maybe it bisects the plots of land. But it definitely isn't perpendicular. Granted, who would visit the perpendicular tower of Pisa? These folks definitely ignored some building codes.

Anyway, perpendicular bisectors come with their very own theorem. The **perpendicular bisector theorem** states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the segment's endpoints. In other words, if we hanged laundry lines from any floor of our tower, each floor would use the same length of laundry line to reach the ground. Okay, but I'm guessing the neighbors might complain about all the underwear hanging outside our tower.

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*.

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.

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!

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*.

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*.

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?

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*.

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.

At the end of this lesson, you should be able to recall the perpendicular bisector theorem and test converse statements.

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