Question

The side $BC$ of a $△ABC$ is produced to ray $BC$ such that $D$ is on ray $BC$. The bisector of $\angle A$ meets $BC$ in $L$. If $\angle ABC+\angle ACD=k\angle ALC$, find $k$.

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is B $2$

Given: The side $BC$ of a triangle $ABC$ is produced to $D$.

The bisector of $\angle A$ meets $BC$ in $L$.

To prove: $\angle ABC+\angle ACD=2\angle ALC$

Proof: In Triangle $ABL$.

Let $\angle BAL=a$ and $\angle LBA=b$

Exterior $\angle ALC=\angle BAL+\angle LBA=a+b$

In triangle $ABC$

As $AL$ bisects $\angle BAC$, $\angle LAC=a$

Exterior $\angle \angle ACD=\angle BAC+\angle ABC=2a+b$

$\angle ABC+\angle ACD=b+(2a+b)$ $2(a+b)=2\angle ALC$

Hence, $\angle ABC+\angle ACD=2\angle ALC$