As per the **Angle Bisector theorem**, the angle bisector of a triangle bisects the opposite side in such a way that the ratio of the two line-segments is proportional to the ratio of the other two sides. Thus the relative lengths of the opposite side (divided by angle bisector) are equated to the lengths of the other two sides of the triangle. This theorem is applicable to all types of triangles.

Class 10 students can read the concept of angle bisector theorem here along with the proof. Apart from angle bisector we will also discuss here the external angle theorem, perpendicular bisector theorem, converse of angle bisector theorem.

## What is Angle Bisector Theorem?

An angle bisector is a straight line drawn from the vertex of a triangle to its opposite side in such a way, that it divides the angle into two equal or congruent angles. Now let us see, what is the angle bisector theorem.

## Interior Angle Bisector Theorem

In the triangle ABC, the angle bisector intersects side BC at point D. See the figure below.

According to the Angle bisector theorem, the ratio of the line segment BD to DC equals the ratio of the length of the side AB to AC.

\(\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |}{\left | AC \right |}\)

Conversely, when a point D on the side BC divides BC in the ratio similar to the sides AC and AB, then the angle bisector of *∠ A *is AD. Hence, according to the theorem, if D lies on the side BC, then,

\(\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |Sin\angle DAB}{\left | AC \right |Sin\angle DAC}\)

If D is external to the side BC, directed angles and directed line segments are required to be applied in the calculation.

Angle bisector theorem is applied when side lengths and angle bisectors are known.

### Proof

We can easily prove the angle bisector theorem, by using trigonometry here. In triangles ABD and ACD (in the above figure) using law of sines, we can write;

\(\frac{AB}{BD}=\frac{sin\angle BDA}{sin\angle BAD}\) ….(1)

\(\frac{AC}{DC}=\frac{sin\angle ADC}{sin\angle DAC}\) ….(2)

The angles *∠ ADC* and *∠ BDA* make a linear pair and hence called adjacent supplementary angles.

Since the sine of supplementary angles are equal, therefore,

Sin ∠ BDA* = *Sin ∠ ADC …..(3)

*Also, *

*∠ DAC* = *∠ BAD *(AD is the angle bisector)

Thus,

Sin ∠ BDA* = *Sin ∠ ADC …(4)

Hence, from equation 3 and 4, we can say, the RHS of equation 1 and 2 are equal, therefore, LHS will also be equal.

\(\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |}{\left | AC \right |}\)

Hence, angle bisector theorem is proved.

**Condition:**

If the angles *∠ DAC* and *∠ BAD* are not equal, the equation 1 and equation 2 can be written as:

\(\frac{\left | AB \right |}{\left | BD \right |}\) sin ∠ BAD = sin∠ BDA

\(\frac{\left | AC \right |}{\left | DC \right |}\) sin ∠ DAC = sin∠ ADC

Angles *∠ ADC* and *∠ BDA* are supplementary, hence the RHS of the equations are still equal. Hence, we get

\(\frac{\left | AB \right |}{\left | BD \right |}\) sin ∠BAD = \(\frac{\left | AC \right |}{\left | DC \right |}\) sin ∠DAC

This rearranges to generalized view of the theorem.

## Converse of Angle Bisector Theorem

In a triangle, if the interior point is equidistant from the two sides of a triangle then that point lies on the angle bisector of the angle formed by the two line segments.

## Triangle Angle Bisector Theorem

Extend the side CA to meet BE to meet at point E, such that BE//AD.

Now we can write,

CD/DB = CA/AE (since AD//BE) —-(1)

∠4 = ∠1 [corresponding angles]

∠1 = ∠2 [AD bisects angle CAB]

∠2 = ∠3 [Alternate interior angles]

∠3 = ∠4 [By transitive property]

ΔABE is an isosceles triangle with AE=AB

Now if we replace AE by AB in equation 1, we get;

CD/DB = CA/AB

Hence proved.

## Perpendicular Bisector Theorem

According to this theorem, if a point is equidistant from the endpoints of a line segment in a triangle, then it is on the perpendicular bisector of the line segment.

Alternatively, we can say, the perpendicular bisector bisects the given line segment into two equal parts, to which it is perpendicular. In case of triangle, if a perpendicular bisector is drawn from the vertex to the opposite side, then it divides the segment into two congruent segments.

In the above figure, the line segment SI is the perpendicular bisector of WM.

## External Angle Bisector Theorem

The external angle bisector of a triangle divides the opposite side externally in the ratio of the sides containing the angle. This condition occurs usually in non-equilateral triangles.

### Proof

Given : In ΔABC, AD is the external bisector of ∠BAC and intersects BC produced at D.

To prove : BD/DC = AB/AC

Constt: Draw CE ∥ DA meeting AB at E

Since, CE ∥ DA and AC is a transversal, therefore,

∠ECA = ∠CAD (alternate angles) ……(1)

Again, CE ∥ DA and BP is a transversal, therefore,

∠CEA = ∠DAP (corresponding angles) —–(2)

But AD is the bisector of ∠CAP,

∠CAD = ∠DAP —–(3)

As we know, Sides opposite to equal angles are equal, therefore,

∠CEA = ∠ECA

In ΔBDA, EC ∥ AD.

BD/DC = BA/AE [By Thales Theorem]

AE = AC,

BD/DC = BA/AC

Hence, proved.

## Solved Examples

Go through the following examples to understand the concept of the angle bisector theorem.

**Example 1:**

Find the value of x for the given triangle using the angle bisector theorem.

**Solution:**

Given that,

AD = 12, AC = 18, BC=24, DB = x

According to angle bisector theorem,

AD/AC = DB/BC

Now substitute the values, we get

12/18 = x/24

X = (⅔)24

x = 2(8)

x= 16

Hence, the value of x is 16.

**Example 2:**

ABCD is a quadrilateral in which the bisectors of angle B and angle D intersects on AC at point E. Show that AB/BC = AD/DC

**Solution:**

From the given figure, the segment DE is the angle bisector of angle D and BE is the internal angle bisector of angle B.

Hence, the using internal angle bisector theorem, we get

AE/EC = AD/DC ….(1)

Similarly,

AE/EC = AB/BC ….(2)

From equations (1) and (2), we get

AB/BC = AD/DC

Hence, AB/BC = AD/DC is proved.

**Example 3**.

In a triangle, AE is the bisector of the exterior ∠CAD that meets BC at E. If the value of AB = 10 cm, AC = 6 cm and BC = 12 cm, find the value of CE.

Solution:

Given : AB = 10 cm, AC = 6 cm and BC = 12 cm

Let CE is equal to x.

By exterior angle bisector theorem, we know that,

BE / CE = AB / AC

(12 + x) / x = 10 / 6

6( 12 + x ) = 10 x [ by cross multiplication]

72 + 6x = 10x

72 = 10x – 6x

72 = 4x

x = 72/4

x = 18

CE = 18 cm

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