# In an Isoceles triangle ABC,bisector CD of the angle C is equal to the base BC.Then the angle between CDA is ?

Jul 2, 2018

$m \angle C D A = {108}^{o}$

#### Explanation:

Since $A B C$ is isosceles with base $B C$, we know that $A B = A C$. Therefore, $m \angle A B C = m \angle A C B$. Since we are given that $C D = B C$, we know that $m \angle C D B = m \angle C B D$. Let $x = m \angle D C B$. Since $C D$ is the bisector of $\angle C$, we know that $m \angle B C A = 2 x = m \angle A B C = m \angle C D B$. Since $C D B$ is a triangle, we know that $m \angle C D B + m \angle C B D + m \angle B C D = {108}^{o} = x + 2 x + 2 x$.
$5 x = {180}^{o}$
$\therefore x = {36}^{o}$.
$\therefore m \angle C D B = 2 x = {72}^{o}$
Since $\angle C D B$ and $\angle C D A$ are supplementary, $m \angle C D B + m \angle C D A = {180}^{o}$
$\therefore {72}^{\oplus} m \angle C D A = {180}^{o}$
$\therefore m \angle C D A = {108}^{o}$