Gfilui

In the figure, in between triangles $ABD$ and $ADC$ the angles $BAD$ and $DAC$ are equal. So are the opposite sides equal i.e. $BD = CD$?

Not necessarily.

Consider for example, this triangle where the angles are exactly the same as in yours but BD is much longer than CD:

Not necessarily. All we can say is that $BD$ is the bisector of the angle $BAC$. If $BD$ is a height then we can say that $BD=DC$ as the triangles $ABD$ and $ADC$ will be congruent (common side $AD$ and two angles).

Not always, because referring to the Angle Bisector Theorem:
$$fracABAC=fracBDCD.$$
The condition $BD=CD$ implies (requires) $AB=AC$.

Not necessarily.

All that we have in this case is a common side and an equal angle which is not enough to conclude that the triangles are equal. You need more information about the triangles. For example if AD is Perpendicular to BC then you have enough information to conclude $BD=CD$

No, it's not true unless AD is perpendicular to the side BC
for your understanding here are some counterexamples.

as you can see in the first triangle it's clear that those two sides are not equal also see the other example where I have taken the angle ACB=90