Question

State and prove interior angle bisector theorem.

Answer 1

Hint: We know that a line segment that bisects one of the vertex angles of a triangle is called the angle bisector of triangle. This angle bisector of an angle of a triangle divides the opposite side into two segments. The interior bisector theorem gives the relation between these two segments and other two sides of the triangle.

Complete step-by-step answer:
Interior angle bisector theorem states that the angle bisector of an angle of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
Now, we will prove the interior angle bisector theorem.
Given: In $△ABC$ , $\overline{BD}$ bisects $\mathrm{\angle }ABC$ .

To prove: ${\displaystyle \frac{AD}{CD}}={\displaystyle \frac{AB}{CB}}$
Proof:

It is given that in $△ABC$ , $\overline{BD}$ bisects $\mathrm{\angle }ABC$ .
We know that an angle bisector is a ray in the interior of an angle which forms two congruent angles.
 $\Rightarrow \mathrm{\angle }ABD\cong \mathrm{\angle }CBD$
Now, we will draw an auxiliary line through $A$ parallel to bisector $\overline{BD}$ which intersects the extension of $\overline{CB}$ at point $E$ as shown in figure.
If two parallel lines are cut by a transversal, the corresponding angles are congruent angles.
 $\Rightarrow \mathrm{\angle }CBD\cong \mathrm{\angle }AEB$
If two parallel lines are cut by a transversal, the alternate interior angles are congruent angles.
 $\Rightarrow \mathrm{\angle }CBD\cong \mathrm{\angle }BAE$
Now, substitute $\mathrm{\angle }CBD\cong \mathrm{\angle }AEB$
 $\Rightarrow \mathrm{\angle }AEB\cong \mathrm{\angle }BAE$
We know from the Side Splitter Theorem that If a line is parallel to one side of a triangle and intersects the other two sides, it divides the sides proportionally.
 $\Rightarrow {\displaystyle \frac{AD}{CD}}={\displaystyle \frac{EB}{CB}}$
The sides opposite the angles are congruent if two angles of a triangle are congruent.
$$\Rightarrow \overline{AB}=\overline{EB}$$
Also, congruent segments have equal lengths.
$$\Rightarrow AB=EB$$
Substituting this in the equation ${\displaystyle \frac{AD}{CD}}={\displaystyle \frac{EB}{CB}}$
 $\Rightarrow {\displaystyle \frac{AD}{CD}}={\displaystyle \frac{AB}{CB}}$
Hence, the theorem is proved.

Note: We can understand the application of the interior angle bisector by solving one simple example.
Suppose we are asked to find the value of $x$ in the given figure where $\overline{BD}$ bisects $\mathrm{\angle }ABC$ in $△ABC$ .

We will apply the interior angle bisector theorem here, then
$$\begin{gathered} {\displaystyle \frac{AD}{CD}}={\displaystyle \frac{AB}{CB}} \\ \Rightarrow {\displaystyle \frac{9}{15}}={\displaystyle \frac{x}{20}} \\ \Rightarrow x={\displaystyle \frac{9\times 20}{15}} \\ \Rightarrow x=12 \end{gathered}$$
Thus, we can find the missing value of a segment by applying this theorem.