### If $ABC$ is a triangle and $D$ is a point on side $AB$ with $AD = BD = CD$, find the value of $\angle ACB$.

$90^\circ$

Step by Step Explanation:
1. It is given that $D$ is a point on the side $AB$ of a $\triangle ABC$ such that $AD = BD = CD$.

We are required to find the value of $\angle ACB.$
2. We are given,
\begin{align} &AD = CD \\ \implies & \angle DAC = \angle DCA \text{ (Angles opposite to equal sides of a triangle) } && \ldots(1) \space\space\space\space\space\space \end{align}
Also,
\begin{align} &BD = CD \\ \implies & \angle DBC = \angle DCB \text{ (Angles opposite to equal sides of a triangle) } && \ldots(2) \space\space\space\space\space\space \end{align}
3. In $\triangle ABC$,
\begin{align} & \angle BAC + \angle ACB + \angle CBA = 180^{ \circ } && \text{ [Angle sum property of a Triangle]} \\ \implies & \angle DAC + \angle ACB + \angle DBC = 180^{ \circ } \\ \implies & \angle DCA + \angle ACB + \angle DCB = 180^{ \circ } && \text{ [By eq (1) and (2)]}\\ \implies & \angle ACB + \angle ACB = 180^{ \circ } \\ \implies & 2\angle ACB = 180^{ \circ } \\ \implies & \angle ACB = 90^{ \circ } \\ \end{align}
4. Hence, the value of $\angle ACB$ is $90^{ \circ }$.