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Geometry: High School15 chapters | 160 lessons

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Lesson Transcript

Instructor:
*Jeff Calareso*

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

The angle bisector theorem sounds almost too good to be true. In this lesson, we set out to prove the theorem and then look at a few examples of how it's used.

It's time to play detective. There's a theorem involving angle bisectors and triangles that sounds a little fishy. Let's do some investigating and see what we can find.

Here's triangle ABC:

This was always a good triangle, never getting into trouble. One day, it got itself mixed up with an angle bisector. That's right - this line from A to BC. We label the point where the angle bisector hits BC as point D. Since it's an angle bisector, it bisects the angle from which it's drawn.

That means angle BAD is congruent to angle CAD. Everything seemed great for the triangle...at first. It was said that there was a theorem we could use - the angle bisector theorem. But it sounded too good to be true. The **angle bisector theorem** states that an angle bisector divides the opposite side of a triangle into two segments that are proportional to the triangle's other two sides. In other words, AB/BD = AC/CD.

How could that be true? The angle bisector makes two smaller triangles that are proportional to each other. Somebody had to prove the theorem, and that's where we come in.

We all wanted the theorem to be true. But is it? We'll need to get our hands a little dirty to find out. To start, let's extend our angle bisector, AD, out a little further. Now let's add a line that's parallel to AB that hits point C and crosses our extended bisector. We'll label this point F.

We can hardly recognize poor old triangle ABC anymore. It's sad, I know. But this is what the triangle wanted. And, trust me, if we want to prove that AB/BD = AC/CD, we need to break some eggs. Well, by breaking eggs, I mean adding lines and stuff.

Okay, time to start putting the pieces together. We know angle BAD equals angle DFC. Why? If AB and FC are parallel, then these are alternate interior angles, and alternate interior angles are equal. That means that angle DFC also equals angle CAD. Remember, BAD and CAD are equal because of the angle bisector.

If we look at triangle ACF below, we have two equal angles, which makes this an isosceles triangle. So, AC = FC. I told you we'd have to break some eggs to solve this case. And we're not there yet. But we're close.

Let's look at two more angles. Angle ADB is congruent to angle CDF. Why? They're vertical angles. And vertical angles are congruent.

Now look at those two small triangles above - ADB and FDC - where we have two congruent angles. We want to be sure to match the right angles - A to F, D to D and B to C. That means that we can state that triangle ADB is similar to triangle FDC because of the angle-angle similarity. Similar triangles are in proportion to one another. Yep, the dots are all connecting now, aren't they? So we can say that AB/BD = FC/CD.

AB/BD = FC/CD...that looks sort of familiar, doesn't it? If this equation were in a line-up, it'd be like our theorem, but maybe it's wearing a fake mustache. And do you remember what FC equals? AC! So, if we swap it out, we get AB/BD = AC/CD.

Did we just prove our theorem? We did. Are we awesome detectives? Pretty much.

So, that's all the proof we need for this angle bisector theorem. What does it look like in practice? Here's triangle XYZ with angle bisector XS:

Let's say we know that XY is 10 and XZ is 12. If YS is 5, what is ZS?

According to the angle bisector theorem, these sides and segments are in proportion to one another like this: XY/YS = XZ/ZS. Let's just plug in what we know and solve. That's 10/5 = 12/*x*. So, 10*x* = 12 * 5. 12 * 5 is 60. Divide that by 10 to get 6. So, ZS is 6. Thanks, angle bisector theorem!

We can also use the theorem to determine if a line is or isn't an angle bisector. Consider this triangle, MNO:

We know that MO is 21, NO is 28, MP is 15 and NP is 20. Is OP an angle bisector? If it is, then MO/MP = NO/NP. Let's test it. That's 21/15 = 28/20. If we cross-multiply, we have 21 * 20 = 15 * 28.

21 * 20 is 420. And 15 * 28? Also 420. So, OP is an angle bisector. Case closed.

In summary, we did some good detective work here. We looked at the **angle bisector theorem**. This theorem states that an angle bisector divides the opposite side of a triangle into two segments that are proportional to the triangle's other two sides. In the triangle below, that's AB/BD = AC/CD.

We then used the theorem to find the missing length in a triangle with an angle bisector. We also used the theorem to determine if a line in a triangle is or isn't an angle bisector.

After you've completed this lesson, you'll have the ability to:

- Describe the angle bisector theorem
- Summarize how to prove the angle bisector theorem
- Use this theorem to find a missing side length or determine whether a line is an angle bisector

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Geometry: High School15 chapters | 160 lessons

- Applications of Similar Triangles 6:23
- Triangle Congruence Postulates: SAS, ASA & SSS 6:15
- Congruence Proofs: Corresponding Parts of Congruent Triangles 5:19
- Converse of a Statement: Explanation and Example 5:09
- Similarity Transformations in Corresponding Figures 7:28
- How to Prove Relationships in Figures using Congruence & Similarity 5:14
- Practice Proving Relationships using Congruence & Similarity 6:16
- The AAS (Angle-Angle-Side) Theorem: Proof and Examples 6:31
- The HA (Hypotenuse Angle) Theorem: Proof, Explanation, & Examples 5:50
- The HL (Hypotenuse Leg) Theorem: Definition, Proof, & Examples 6:19
- Perpendicular Bisector Theorem: Proof and Example 6:41
- Angle Bisector Theorem: Proof and Example 6:12
- Congruency of Isosceles Triangles: Proving the Theorem 4:51
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