$D$ is any point on side $AC$ of a $\triangle ABC$ with $AB = AC$. Show that $CD < BD$.

(N/A) In $\triangle ABC$,we have $AB = AC$ (Given).
Therefore,$\angle ABC = \angle ACB$ (Angles opposite to equal sides of a triangle are equal).
Since $D$ is a point on $AC$,$\angle DBC < \angle ABC$.
Substituting $\angle ACB$ for $\angle ABC$,we get $\angle DBC < \angle ACB$.
Since $\angle ACB$ is the same as $\angle DCB$,we have $\angle DBC < \angle DCB$.
In $\triangle BCD$,since $\angle DCB > \angle DBC$,the side opposite to the greater angle is longer.
Therefore,$BD > CD$ or $CD < BD$.