To prove that $D$ is the midpoint of $\overline{AB}$ given that $\triangle ABC$ has $\angle ACB = 90^\circ$ and $AD = CD$, we can proceed as follows:

Step-by-Step Proof

1. Identify Triangle Properties:

   • $\triangle ABC$ is a right triangle with $\angle ACB = 90^\circ$.
   • Point $D$ is on side $\overline{AB}$ such that $AD = CD$.

2. Triangles $\triangle ACD$ and $\triangle BCD$:

   • Since $AD = CD$, $\triangle ACD$ is isosceles with $\angle DAC = \angle DCA$.
   • Let these angles each be $x^\circ$. So, $\angle DAC = \angle DCA = x^\circ$.

3. Angle Calculation:

   • In $\triangle ACD$, the sum of angles must be $180^\circ$. Therefore, $\angle DAC + \angle DCA + \angle ACD = x^\circ + x^\circ + 90^\circ = 180^\circ.$
   • Simplifying, $2x^\circ + 90^\circ = 180^\circ$.
   • So, $2x^\circ = 90^\circ$ and $x^\circ = 45^\circ$. Thus, $\angle DAC = \angle DCA = 45^\circ$.

4. Consider Triangle $\triangle BCD$:

   • Since $\angle ACB = 90^\circ$, we have: $\angle BCD = 90^\circ - x^\circ = 90^\circ - 45^\circ = 45^\circ.$
   • Hence, $\angle BCD = 45^\circ$.

5. Identify Isosceles Triangles:

   • In $\triangle BCD$, since $\angle DCB = 45^\circ$ and $\angle BCD = 45^\circ$, $\triangle BCD$ is also isosceles with $BD = CD$.

6. Conclude $D$ is the Midpoint of $\overline{AB}$:

   • Since $AD = CD$ and $BD = CD$, we have $AD = BD$.
   • Therefore, $D$ is equidistant from $A$ and $B$, which means $D$ is the midpoint of $\overline{AB}$.

Summary:

Since $AD = BD$, point $D$ divides $\overline{AB}$ into two equal parts. Hence, $D$ is the midpoint of $\overline{AB}$.

Would you like more details or have any questions? Here are some related problems:

1. How would the proof change if $\angle ACB$ was not $90^\circ$?
2. What properties of isosceles triangles are crucial in this proof?
3. Can you prove that $\triangle ACD$ and $\triangle BCD$ are congruent?
4. How would the coordinates of $D$ relate to $A$ and $B$ if $\triangle ABC$ was placed on a coordinate plane?
5. What is the significance of $\angle ACB = 90^\circ$ in the proof?

Tip: Remember, identifying isosceles triangles can often simplify complex geometric proofs by providing equal angles and sides!