Given:
The circle centered at point
\[A\] has a radius of length
\[AB\].
The circles centered at points
\[B\] and
\[D\] have radii of length
\[DE\].
Ray A B and ray A C intersect at point A. A circle arc centered at point A interesects point B on ray A B and point D on ray A C. One circle arc centered at point B and one circle arc centered at point D intersect at point A and point E. A ray connects points A and E so that rays A B, A D, and A E all start at point A.
\[A\]
\[B\]
\[C\]
\[D\]
\[E\]
Complete the proof that
\[\angle DAE \cong \angle BAE\].
Step	Statement	Reason
1
\[AB=AD\]	All radii of the same circle have the same length.
2

Both circles have radii of the same length.
3
\[AE=AE\]	They're lengths of the same segment.
4

Side-side-side congruence (1, 2, 3)
5
\[\angle DAE \cong \angle BAE\]	Corresponding parts of congruent triangles are congruent (4).

Let's complete the proof by filling in the missing steps in the two-column format. The goal is to prove that \(\angle DAE \cong \angle BAE\).

### Proof:

| **Step** | **Statement** | **Reason** |
|:--|:--|:--|
| 1 | \(AB = AD\) | All radii of the same circle have the same length. |
| 2 | \(BE = DE\) | Both circles centered at \(B\) and \(D\) have the same radius (length \(DE\)). |
| 3 | \(AE = AE\) | Reflexive property (a segment is equal to itself). |
| 4 | \(\triangle ABE \cong \triangle ADE\) | Side-Side-Side (SSS) congruence (steps 1, 2, and 3). |
| 5 | \(\angle DAE \cong \angle BAE\) | Corresponding parts of congruent triangles are congruent (CPCTC) (step 4). |

### Explanation:

- **Step 1:** We know that \(AB\) and \(AD\) are radii of the circle centered at \(A\), so they are equal in length.
- **Step 2:** Circles centered at \(B\) and \(D\) have the same radius, which is \(DE\), so the lengths \(BE\) and \(DE\) are equal.
- **Step 3:** The segment \(AE\) is shared between the two triangles \(\triangle ABE\) and \(\triangle ADE\), so it is trivially equal to itself.
- **Step 4:** Using the SSS postulate, triangles \(\triangle ABE\) and \(\triangle ADE\) are congruent because all corresponding sides are equal.
- **Step 5:** Since \(\triangle ABE \cong \triangle ADE\), the angles \(\angle DAE\) and \(\angle BAE\) are congruent by CPCTC (corresponding parts of congruent triangles).

This completes the proof that \(\angle DAE \cong \angle BAE\).

---

Do you want further details on any of these steps or have any questions?

Here are 5 relative questions to expand on the topic:

1. What is the SSS postulate and how is it used to prove triangle congruence?
2. How does the CPCTC rule apply when proving angle congruence in triangles?
3. What other methods can be used to prove triangles congruent besides SSS?
4. Why is the reflexive property important in geometric proofs?
5. How would you prove that two angles are congruent without using triangle congruence?

**Tip:** In geometry, always look for congruent triangles as a powerful tool for proving equal angles, sides, and other properties!

Given: The circle centered at point A has a radius of length AB. The circles centered at points B and D have radii of length DE. Ray AB and ray AC intersect at point A. A circle arc centered at point A intersects point B on ray AB and point D on ray AC. One circle arc centered at point B and one circle arc centered at point D intersect at point A and point E. A ray connects points A and E so that rays AB, AD, and AE all start at point A. Complete the proof that ∠DAE ≅ ∠BAE.

Explore the step-by-step geometric proof that ∠DAE ≅ ∠BAE. This proof uses the SSS triangle congruence theorem and the CPCTC rule, which demonstrates the congruence of corresponding angles in congruent triangles. Suitable for students in Grades 9-12.

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