An

**angle bisector** is defined as the line which cuts the angle into equal angles. If any segment, ray or light passing through the vertex of the angle divides the angles into two congruent angles then it is called the

**angle bisector.****Angle Bisector Theorems:**

The angle bisector theorem states that, in given triangle say $PQR$ and angle bisector $PS$, where $S$ lies on the side $PQ$, then $\frac{q}{m}$ = $\frac{r}{n}$

**Prove:**

We need to two similar triangles to prove the angle bisector theorem. We see that the given triangle $PQR$ is separated into two smaller triangles** $PRS$ and $PQS$** which are not similar to each other. So, we need to draw another triangle which shall be a similar triangle.

To do that, we extend the angle bisector $PS$ and draw a line parallel to $PR$ originating from the vertex $Q$ and intersecting the extended angle bisector

**$PS$ at $T$ (say).**
Let us take up the triangles $PRS$ and $TQS$,

Angle $RPS$ = Angle $QTS$ (Alternate Interior angles)

Angle $RSP$ = Angle $QST$ (Vertically opposite angles)

Therefore, Triangle $PRS$ similar to Triangle $TQS$ (AA postulate)

So, their corresponding sides are in proportion:

$\frac{PR}{RS}$ = $\frac{TQ}{QS}$

$\frac{q}{m}$ = $\frac{u}{n}$ ...... **Equation 1**

Now, we see that angle $QPS$ = angle $RPS$ (angle bisector)

Angle $QPS$ = Angle $QTS$ (As angle $RPS$ and angle $QTS$ are alternate interior angles)

Let us consider the bigger triangle $PQT$, where angle $QPS$ and angle $QTS$ are the base angles. As the base angles are equal their corresponding sides are also equal forming an isosceles triangle $PQT$. So, $PQ$ = $TQ$ or $r$ = $u$. Plugging in this value of $r$ = $u$ in equation $1$, we get $\frac{q}{m}$ = $\frac{r}{n}$. Hence, the angle bisector theorem is proved.