# Angle Bisector Theorem

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## What is Angle Bisector Theorem?

Angle Bisector Theorem states that "In a triangle, the angle bisector of any angle will divide the opposite side in the ratio of the sides containing the angle".

Consider the figure below, in which $$AD$$ is the bisector of $$\angle A$$. Angle Bisector Theorem tells us that $$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$

## Proof of Angle Bisector Theorem:

Consider $$\Delta ABC$$,

Given: $$AD$$ is the bisector of $$\angle A$$

To Prove: $$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$

Construction: Draw a ray $$CX$$ parallel to $$AD$$, and extend $$BA$$ to intersect this ray in $$E$$ Proof:

By the Basic Proportionality Theorem, We know that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio

$$\therefore$$ In $$\Delta CBE$$, $$DA\parallel CE$$

${\text{}} \Rightarrow \frac{{BD}}{{DC}} = \frac{{BA}}{{AE}}....(1)$

Now, we only need to prove that $$AE = AC$$.

For that, consider the figure above once again, with the angles marked, as shown below: Since $$DA\parallel CE$$, We have:

• $$\angle1$$ = $$\angle 2$$ (corresponding angles)

• $$\angle 3$$ = $$\angle 4$$ (alternate interior angles)

But $$AD$$ is the bisector of $$\angle BAC$$, which means that $$\angle 1$$ = $$\angle 3$$.

Thus, $$\angle 2$$ = $$\angle 4$$.

$\Rightarrow AC = AE{\text{ }}({\text{Sides opposite to equal angles are equal}})....{\text{(2)}}$

Finally, from (1) and (2),

$\boxed{\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}}$

The angle bisector $$AD$$ thus divides the opposite side $$BC$$ in the ratio of the sides $$(AB$$ and $$AC)$$ containing that angle.

Hence Proved.

✍Note: Angle bisector theorem is an important consequence of BPT. Challenge: In $$\Delta PQR$$, $$E$$ is a point on $$QR$$ such that $$PE$$ is the bisector of $$\angle P$$

$$PQ = 4$$ units, $$QE = 2$$ units, $$ER = 3$$ units.

Find $$PR$$.

⚡Tip: By Angle Bisector Theorem, $$\frac{{QE}}{{ER}} = \frac{{PQ}}{{PR}}$$