Question

In $△$ABC, AC $>$ AB and AD bisects $\angle$A. Which of the following option is correct?

• A. $\angle ADC$ $>$ $\angle ADB$
• B. $\angle ADC$ = $\angle ADB$
• C. $\angle ADC$ $<$ $\angle ADB$
• D. Can't say

Answer 1

The correct option is A

$\angle ADC$ $>$ $\angle ADB$

In $△ABC$, it is given that AC $>$ AB, which means $\angle B$ $>$ $\angle C$.

Now, AD bisects $\angle A$,
so, in $△ADC$, $\angle ADC$ = ${180}^{\circ }$ - $(\frac{\angle A}{2}$ + $\angle C)$
Similarly, in $△ADB$, $\angle ADB$ = ${180}^{\circ }$ – $(\frac{\angle A}{2}$ + $\angle B)$

and since $\angle B$ in $△ADB$ is greater than $\angle C$ in $△ADC$, and $\frac{\angle A}{2}$ is common (AD is the bisector),
we can deduce that $\angle ADB$ is smaller than $\angle ADC$, and the side opposite to the bigger angle would be larger than the other.