In a , = 50^ and BC is produced to a point D. If the bisectors meet at E, then =

In a , = 50^ and BC is produced to a point D. If the bisectors me…

MCQ

In a $\triangle\text{ABC}, \ \angle\text{A} = 50^\circ$ and $BC$ is produced to a point $D$. If the bisectors $\angle\text{ACD}$ meet at $E$, then $\angle\text{E} =$

✓
$25^\circ$
B
$50^\circ$
C
$75^\circ$
D
$100^\circ$

✓

Answer

Correct option: A.

$25^\circ$

$BE$ and $CE$ are bisectors of $\angle\text{B}$ and $\angle\text{ACD}.$
in $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{B}+\angle\text{C}=180^\circ-50^\circ=130^\circ\ ...\ \text{(i)}$
Now,in $\triangle\text{BEC}$
$\angle\text{CBE}+\angle\text{BEC}+\angle\text{ECB}=180^\circ$
$\angle\text{CBE}=\frac{\angle\text{B}}{2},\ \angle\text{BEC}=\angle\text{E},\ \angle\text{ECB}=\angle\text{C}+\angle\text{ACE}$
Now, $\angle\text{ACD}=180^\circ-\angle\text{C}$
$\angle\text{ACE}=\frac{\angle\text{ACD}}{2}=\frac{180^\circ-\angle\text{C}}{2}=90^\circ-\frac{ \angle\text{C}}{2}$
So, $\angle\text{ECB}=\angle\text{C}+90^\circ-\frac{\angle\text{C}}{2}$
$\Rightarrow\ \angle\text{ECB}=90^\circ+\frac{\angle\text{C}}{2}$
Now putting all values in eq. $(ii)$
$\frac{\angle\text{B}}{2}+\angle\text{E}+90^\circ+\frac{\angle\text{C}}{2}=180^\circ$
$\Rightarrow\ \angle\text{E}=180^\circ-90^\circ-\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$
$=90^\circ-\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$
$=90^\circ-\frac{130^\circ}{2}$
$\Rightarrow\ \angle\text{E}=25^\circ$