- Definitions
- Perpendicular Bisector Theorem
- Proving the Theorem
- Practice Proof
- Converse of the Theorem

All good learning begins with vocabulary, so we will focus on the two important words of the theorem. **Perpendicular** means two line segments, rays, lines or any combination of those that meet at right angles. A line is perpendicular if it intersects another line and creates right angles.

A **bisector** is an object (a line, a ray, or line segment) that cuts another object (an angle, a line segment) into two equal parts. A bisector cannot bisect a line, because by definition a line is infinite.

Putting the two meanings together, we get the concept of a **perpendicular bisector**, a line, ray or line segment that bisects an angle or line segment at a right angle.

Before you get all bothered about it being a perpendicular bisector of an *angle*, consider: what is the measure of a straight angle? $180\xb0$; that means a line dividing that angle into two equal parts and forming two right angles is a perpendicular bisector of the angle.

Okay, we laid the groundwork. So putting everything together, what does the **Perpendicular Bisector Theorem** say?

If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment.

Suppose you have a big, square plot of land, $\mathrm{1,000}meters$ on a side. You built a humdinger of a radio tower, $300meters$ high, right smack in the middle of your land. You plan to broadcast rock music day and night.

Anyway, that location for your radio tower means you have $500meters$ of land to the left, and $500meters$ of land to the right. Your radio tower is a perpendicular bisector of the length of your land.

You need to reinforce the tower with wires to keep it from tipping over in high winds. Those are called guy wires. How long should a guy wire from the top down to the land be, on each side?

Because you constructed a perpendicular bisector, you do not need to measure on each side. One measurement, which you can calculate using geometry, is enough. Use the **Pythagorean Theorem** for right triangles:

${a}^{2}+{b}^{2}={c}^{2}$

Your tower is $300meters$. You can go out $500meters$ to anchor the wire's end. The tower meets your land at $90\xb0$. So:

$300{m}^{2}+500{m}^{2}={c}^{2}$

$\mathrm{90,000}+\mathrm{250,000}={c}^{2}$

$\mathrm{340,000}={c}^{2}$

$\sqrt{\mathrm{340,000}}=c$

$583.095m=c$

You need guy wires a whopping $583.095meters$ long to run from the top of the tower to the edge of your land. You repeat the operation at the $200meter$ height, and the $100meter$ height.

For every height you choose, you will cut guy wires of identical lengths for the left and right side of your radio tower, because the tower is the perpendicular bisector of your land.

Behold the awesome power of the two words, "perpendicular bisector," because with only a line segment, $HM$, and its perpendicular bisector, $WA$, we can prove this theorem.

We are given line segment $HM$ and we have bisected it (divided it exactly in two) by a line $WA$. That line bisected $HM$ at $90\xb0$ because it is a given. This means, if we run a line segment from $PointW$ to $PointH$, we can create right triangle $WHA$, and another line segment $WM$ creates right triangle $WAM$.

What do we have now? We have two right triangles, $WHA$ and $WAM$, sharing side $WA$, with all these congruences:

- $WA\cong WA$ (by the reflexive property)
- $\angle WAH\cong \angle WAM$ (90° angles; given)
- $HA\cong AM$ (bisector; given)

What does that look like? We hope you said **Side Angle Side**, because that is exactly what it is.

The **Side Angle Side Postulate** states, "If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent."

That means sides $WH$ and $WM$ are congruent, because CPCTC (corresponding parts of congruent triangles are congruent). $WHAM!$ Proven!

You can tackle the theorem yourself now. You will either sink or swim on this one. Here is a line segment, $WM$. We construct a perpendicular bisector, $SI$.

**How can you prove that $SW\cong SM$? Do you know what to do?**

- Construct line segments $SW$ and $SM$.
- You now have what? Two right triangles, $SWI$ and $SIM$. They have right angles, $\angle SIW$ and $\angle SIM$.
- Identify $WI$ and $IM$ as congruent, because they are the two parts of line segment $WM$ that were bisected by $SI$.
- Identify $SI$ as congruent to itself (by the reflexive property).

What does that give you? Two congruent sides and an included angle, which is what postulate? The **SAS Postulate**, of course! Therefore, line segment $SW\cong SM$.

So, did you sink or $SWIM$?

Notice that the theorem is constructed as an "if, then" statement. That immediately suggests you can write the converse of it, by switching the parts:

If a point is equidistant from the endpoints of a line segment, then it is on the perpendicular bisector of the line segment.

We can show this, too. Construct a line segment $HD$. Place a random point above it (but still somewhere between $PointsH$ and $D$) and call it $PointT$. If $PointT$ is the same distance from $PointsH$ and $D$, this converse statement says it must lie on the perpendicular bisector of $HD$.

You can prove or disprove this by dropping a perpendicular line from $PointT$ through line segment $HD$. Where your perpendicular line crosses $HD$, call it $PointU$.

If $PointT$ ** is** the same distance from $PointsH$ and $D$, then $HU\cong UD$. If $PointT$ is

The symbol $\ncong $ means "not congruent to."

You can go through the steps of creating two right triangles, $\u25b3THU$ and $\u25b3TUD$ and proving angles and sides congruent (or not congruent), the same as with the original theorem.

You would identify the right angles, the congruent sides along the original line segment $HD$, and the reflexive congruent side $TU$. When you got to a pair of corresponding sides that were not congruent, then you would know $PointT$ was not on the perpendicular bisector.

Only points lying on the perpendicular bisector will be equidistant from the endpoints of the line segment. Everything else lands with a $THUD$.

After you worked your way through all the angles, proofs and multimedia, you are now able to recall the Perpendicular Bisector Theorem and test the converse of the Theorem. You also got a refresher in what "perpendicular," "bisector," and "converse" mean.

After studying this work, you will be able to:

- Recall the Perpendicular Bisector Theorem
- Test the converse of the Theorem

Instructor: **Malcolm M.**

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

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