Question

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

• A. $\angle ADC$ $>$ $\angle ADB$
• B. $\angle ADC$ $<$ $\angle ADB$
• C. $\angle ADC$ = $\angle DAB$
• D. $\angle ADC$ = $\angle ADB$

Answer 1

The correct option is A $\angle ADC$ $>$ $\angle ADB$


Given, in $\Delta ABC$, AC $>$ AB.
$\Rightarrow \angle B$ $>$ $\angle C$ (If two sides of a triangle are unequal, the angle opposite to the longer side is larger)
Also it is given that AD bisects $\angle A$.
$\Rightarrow \angle BAD$ = $\angle CAD$ = $\frac{\angle A}{2}$

Sum of angles in a triangle is ${180}^{\circ }$.
So, in $\Delta ADC$,
$\angle ADC$ = ${180}^{\circ }$ - $(\frac{\angle A}{2}$ + $\angle C)$ $\cdots \left(1\right)$

Similarly, in $\Delta ADB$,
$\angle ADB$ = ${180}^{\circ }$ – $(\frac{\angle A}{2}$ + $\angle B)$ $\cdots \left(2\right)$

$\frac{\angle A}{2}$ is common for both the triangles and $\angle B$ in $\Delta ADB$ is greater than $\angle C$ in $\Delta ADC$
We can deduce from (1) and (2) that $\angle ADB$ is smaller than $\angle ADC$.
i.e., $\angle ADC$ $>$ $\angle ADB$.

[Hence the option $\angle ADC$ $<$ $\angle ADB$ is not correct.
Also since the measures of sides and angles are not mentioned in the question, nothing can be deduced with respect to the remaining options.]