Question

In the given figure, the side BC of $\Delta$ ABC is produced to D. If the bisecter of $\angle BAC$ meets BC at L,, then $\angle ABC+\angle ACD$ = ___

• A. $\angle ALC$
• B. $2\angle ALC$
• C. $3\angle ALC$
• D. $4\angle ALC$

Answer 1

The correct option is A

$\angle ALC$

(i) $\angle BAL=\angle LAC=\frac{\angle A}{2}$... (given AL bisects $\angle A)$

(ii) $\angle ALC=\frac{A}{2}+B......$ (exteriar angle theorem)

(iii) $\angle ACD=A+B......$ (exteriar angle theorem)

(iv) $\angle ABC+\angle ACD=B+A+B$

= A + 2B

= 2 $\left(\frac{A}{2}\right)+2B=2(\frac{A}{2}+B)=2\angle ALC$