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
.

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

### Step-by-Step Proof:

#### Step 1: Given Information and Triangle Setup

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

Since $ AD = CD $, triangle $ ACD $ is isosceles, and the two base angles, $ \angle DAC $ and $ \angle ACD $, are congruent.

#### Step 2: Define Angle Measures

Let $ \angle DAC = \angle ACD = x^\circ $, since triangle $ ACD $ is isosceles. Now, because the angles in a triangle sum to $ 180^\circ $, we can find $ \angle DCB $.

In $ \triangle ACB $, we already know that $ \angle ACB = 90^\circ $, so:

\[
\angle ACB + \angle DCB = 90^\circ.
\]

Since the angles at point $ C $ must sum to $ 90^\circ $, we find that:

\[
\angle DCB = 90^\circ - x^\circ.
\]

#### Step 3: Analyze Triangle $ ABC $

Now, in triangle $ ABC $, we can analyze the angle $ \angle BAC $. Since $ \angle ACB = 90^\circ $, the sum of angles in triangle $ ABC $ gives:

\[
\angle ABC + \angle BAC + \angle ACB = 180^\circ.
\]

Substituting known values:

\[
\angle ABC + \angle BAC + 90^\circ = 180^\circ.
\]

Thus,

\[
\angle ABC + \angle BAC = 90^\circ.
\]

Now, from Step 2, we know that $ \angle DCB = 90^\circ - x^\circ $. Therefore, $ \angle ABC = x^\circ $, meaning that:

\[
\angle BAC = 90^\circ - x^\circ.
\]

#### Step 4: Prove $ D $ is the Midpoint

At this point, we have established that triangle $ ABC $ is also isosceles because:

\[
\angle BAC = \angle ABC = 90^\circ - x^\circ.
\]

Since $ \triangle ABC $ is isosceles and both angles $ \angle BAC $ and $ \angle ABC $ are equal, it follows that side $ AB $ is bisected by point $ D $, making $ D $ the midpoint of $ \overline{AB} $.

Thus, $ AD = BD = CD $, and we conclude that $ D $ is indeed the midpoint of $ \overline{AB} $.

### Conclusion:

Point $ D $ is the midpoint of side $ \overline{AB} $, as we have shown that triangle $ ABC $ is isosceles and $ D $ bisects side $ \overline{AB} $.

---

Would you like further clarification or details on any part of the proof? Here are 5 related questions to deepen understanding:

1. How can we verify the midpoint of a line segment using coordinates?
2. What are other properties of isosceles triangles that apply to this problem?
3. How would this proof change if triangle $ ABC $ were not a right triangle?
4. What are the implications of triangle symmetry in problems like this?
5. Can the perpendicular bisector theorem provide an alternative proof?

**Tip:** In geometry, using both algebraic and geometric approaches can often provide deeper insight into a problem!

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

Explore the geometric proof that shows D is the midpoint of side AB in triangle ABC where ∠ACB = 90° and AD = CD. This proof covers key geometric concepts including isosceles triangles and the angle sum theorem, ideal for students in Grades 8-10.

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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