## Key Concepts

- Angle Bisector Theorem
- Concurrency
- In-center

### Introduction

In this session, you will learn about the angle bisector theorem, its converse, the concurrency of angle bisectors of a triangle, and the incenter.

What is an angle bisector?

What are segments?

### Angle Bisector Theorem

#### Angle Bisector

A ray that divides an angle into two adjacent congruent angles is called the angle bisector.

#### Distance from a Point to Line

The length of the perpendicular segment from the point to the line is called the distance from a point to the line.

#### Angle Bisector Theorem

If a point is on the angle bisector, then it is equidistant from the two sides of the angle.

If OC is the angle bisector of ∠AOB, then CA ⊥ OA and CB ⊥ OB,

Then CA = CB

### Converse of Angle Bisector Theorem

#### Theorem

If a point is in the interior of an angle and is equidistant from the two sides of the angle, then the point lies on the angle bisector.

If CA ⊥ OA and CB ⊥ OB, CA = CB, then C lies on angle bisector of ∠AOB

**Example 1:**

Find ∠DAC in the given figure if CD = BD = 10 units.

**Solution:**

If CD ⊥ AC and DB ⊥ AB, DC = DB

D lies on angle bisector of ∠CAB

By the converse of angle bisector theorem

∠DAC= ∠DAB

= 52°

**Example 2:**

In the figure ABCD, if D bisects ∠BAC, find CD.

**Solution:**

D bisects ∠BAC, and

D is equidistant from the sides of the angle

So, CD = DB

Since by using an angle bisector theorem.

CD = DB

3x + 5 = 2x + 10

3x – 2x = 10 – 5

x = 5

CD = 3x + 5

= 3*5 + 5

= 15 + 5

= 20 units

Therefore, D lies on the bisector of ∠BAC, if CD = 20 units.

### Concurrency of Angle Bisectors of a Triangle

#### Theorem

The angle bisectors of a triangle intersect at a common point that is equidistant from the sides of a triangle.

If AP, BP, CP are angle bisectors of the triangle, then XP = YP = ZP

**Example:**

In the figure, if P is the incenter, then find PY.

**Solution:**

Here, we use the point of concurrency for angle bisectors of a triangle theorem.

From that theorem, P is equidistant from the three sides of triangle ABC, so XP = YP = ZP

We must find YP, for that we can find XP from ∆ XBP,

BX = 5 units, BP = 13 units

As, ∆ XBP is a right-angled triangle, we use Pythagorean theorem,

BP^{2} = XP^{2} + BX^{2}

13^{2} = XP^{2} + 5^{2}

13^{2}−5^{2} = XP^{2}

169−25 = XP^{2}

XP^{2} = 144

√XP = 144

XP = 12

### Incenter

The angle bisectors of a triangle intersect at a common point that is equidistant from the sides of the triangle, this point of concurrency is called the incenter of the triangle.

Incenter always lies inside the triangle, the circle drawn with incenter as the center is inscribed in the triangle and it is called incircle.

### Angle Bisectors in Real Life

Angle bisectors in real life are used in architecture, for making quilts and in preparing cakes.

Angle bisectors are used in playing several games as well like rugby, football, billiards.

#### Real Life Example

In Snooker game, the position of the cue is relative to the ball and goal pocket forms congruent angles, as shown.

Will the Snooker player have to move the cue farther to shoot toward the right goal pocket or the left goal socket?

**Solution:**

The congruent angles tell that the cue is placed on the angle bisector of the balls.

By the angle bisector theorem, the snooker player’s cue is equidistant from both left goal pocket and the right goal pocket.

So, the Snooker player must move the same distance to shoot another goal.

## Exercise

- From the given figure, find the relation between AD and ∠BAC.

- From the given figure, if BD ⊥ AB, DC ⊥ AC, then find the relation between BD and DC.

- From the given figure, find relation between P and ∠NOM.

- If P is equidistant from all the sides of the triangle ABC, a circle drawn with P as center touching all the sides, then the center of the circle is also called ______.
- In a scalene triangle, the incenter lies ____.
- Prove the converse of the angle bisector theorem.
- Can you do an activity to show the location of the incenter?
- Identify the theorem used to solve the problem.

- Find ∠DAC in the given figure.

- Find MP in the given figure.

### Concept Map

### What we have learned

- Angle bisector theorem.
- Converse of angle bisector theorem.
- Concurrency.
- Incentre

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