Triangle $ABC$ has $\angle ACB = 90^{\circ}$. Point $D$ is on side $\overline{AB}$ such that $AD = CD$. Prove that $D$ is the midpoint of $\overline{AB}$.
[asy]
pair A, B, C, D;
A = (-5,0);
D = (0,0);
B = (5,0);
C = (-3,4);
draw(A--B--C--cycle);
draw(C--D, dotted);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, S);
[/asy]
Since 
A
D
=
C
D
, 
 is isosceles, giving the equal angles 
=
; let the measure of these angles be 
x
∘
. Then, 
∠
D
C
B
=
−
=
90
∘
−
x
∘
. Since the angles of 
 add to 
180
∘
, 
∠
A
B
C
+
∠
A
C
B
+
∠
B
A
C
=
+
90
∘
+
x
∘
=
180
∘
,
 and solving gives 
=
90
∘
−
x
∘
. Therefore, 
=
, so 
 is isosceles. Therefore, 
B
D
=
C
D
=
A
D
, and we conclude that 
D
 is the midpoint of 
¯¯¯¯¯¯¯¯
A
B
.
∠
A
C
B
∠
D
A
C
∠
D
B
C
∠
D
C
A
∠
D
C
B
△
A
D
C
△
C
D
B
△
A
B
C
∠
C
D
A
∠
C
D
B


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 $\overline{AB}$.

---

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.

Triangle ABC has ∠ACB = 90°. Point D is on side AB such that AD = CD. Prove that D is the midpoint of AB.

Explore how to prove that point D is the midpoint of AB in triangle ABC, where ∠ACB = 90°. This problem uses key geometric concepts such as the Isosceles Triangle Theorem and the Triangle Angle Sum Theorem. Suitable for students in Grades 9-12.

Math Problem Statement

C D , is isosceles, giving the equal angles

; let the measure of these angles be x ∘ . Then, ∠ D C B

−

90 ∘ − x ∘ . Since the angles of add to 180 ∘ , ∠ A B C + ∠ A C B + ∠ B A C

90 ∘ + x ∘

180 ∘ , and solving gives

90 ∘ − x ∘ . Therefore,

, so is isosceles. Therefore, B D

C D

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