We are given that triangle $ABC$ is a right triangle with $\angle ACB = 90^\circ$, and that point $D$ is on side $\overline{AB}$ such that $AD = CD$. We are tasked with proving that $D$ is the midpoint of $\overline{AB}$.

Step 1: Identify the properties of the triangle

Since $AD = CD$, triangle $ADC$ is isosceles, meaning that the two angles opposite the equal sides $AD$ and $CD$ are also equal. Thus, we have:

$\angle DAC = \angle DCA.$

Let $\angle DAC = \angle DCA = x^\circ$. We are also given that $\angle ACB = 90^\circ$, and from the triangle angle sum property, we know that the sum of the angles in any triangle is $180^\circ$. Applying this to triangle $ACB$, we get:

$\angle BAC + \angle ABC + \angle ACB = 180^\circ.$

Substituting $\angle ACB = 90^\circ$ gives:

$\angle BAC + \angle ABC + 90^\circ = 180^\circ,$ which simplifies to:

$\angle BAC + \angle ABC = 90^\circ.$

Step 2: Analyze triangle $ACD$

In triangle $ACD$, we know that $\angle DAC = x^\circ$ and $\angle DCA = x^\circ$. Using the triangle angle sum property again, we find $\angle ACD$:

$\angle ACD + \angle DAC + \angle DCA = 180^\circ.$ Substituting $\angle DAC = x^\circ$ and $\angle DCA = x^\circ$ into the equation:

$\angle ACD + 2x = 180^\circ.$

Since $\angle ACD = 90^\circ$ (as given in the problem), we have:

$90^\circ + 2x = 180^\circ,$ which simplifies to:

$2x = 90^\circ \quad \text{or} \quad x = 45^\circ.$

Thus, $\angle DAC = \angle DCA = 45^\circ$.

Step 3: Analyze triangle $BDC$

Now, let’s consider triangle $BDC$. Since $\overline{AB}$ is a straight line and point $D$ lies on $\overline{AB}$, we know that $\angle BDC = \angle ADC = 45^\circ$, because triangle $ADC$ is isosceles with $\angle DAC = \angle DCA = 45^\circ$.

Step 4: Use symmetry to conclude that $D$ is the midpoint

Since triangle $ADC$ is isosceles with $AD = CD$, and triangle $BDC$ is isosceles with $BD = CD$, we can conclude that $AD = BD$. This means that $D$ is equidistant from points $A$ and $B$, which implies that $D$ is the midpoint of $\overline{AB}$.

Conclusion

We have proven that point $D$ is the midpoint of $\overline{AB}$.

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Would you like more details on any of the steps or clarifications? Here are some related questions for further exploration:

1. How can the midpoint formula be applied in coordinate geometry to verify $D$ as the midpoint?
2. What are the properties of an isosceles triangle and how can they be applied in other problems?
3. How does the right-angle property help simplify angle sums in geometric proofs?
4. Can this proof be adapted using trigonometry instead of purely geometric reasoning?
5. How would the proof change if $\triangle ABC$ were not a right triangle?

Tip: For problems involving isosceles triangles, always check if equal sides give rise to equal angles, and use that symmetry to simplify your proof.