Ans: (b) $42^\circ$
- In $\Delta ABD$ and $\Delta ACD$,
$\because AD$ is the angle bisector of $\angle BAC$,
$\therefore \angle BAD = \angle CAD$.
$\angle ABD = \angle ACD$ (Given)
$AD = AD$ (common side)
$\therefore \Delta ABD \cong \Delta ACD$ (A.A.S.)
- By the result of (a), $\Delta ABD \cong \Delta ACD$. Hence, we have
$\angle CAD = \angle BAD = 31^\circ$ and $\angle ABD = \angle ACD = 17^\circ$ and $BD = CD$.
Since $BD=CD$, $\angle CBD = \angle BCD$.
Hence, we have
$\begin{array}{rcl}
\angle CBD & = & \dfrac{1}{2} \times (180^\circ – \angle ABD – \angle BAD – \angle CAD – \angle ACD) \\
& = & \dfrac{1}{2}(180^\circ – 31^\circ -31^\circ-17^\circ-17^\circ) \\
& = & 42^\circ
\end{array}$
