To solve the problem, we need to show that if \( e^{iA}, e^{iB}, e^{iC} \) are in arithmetic progression (A.P.), where \( A, B, C \) are the angles of a triangle, then the triangle is equilateral.
### Step-by-Step Solution:
1. **Understanding the Condition of A.P.:**
Since \( e^{iA}, e^{iB}, e^{iC} \) are in A.P., we can express this condition mathematically:
\[
2 e^{iB} = e^{iA} + e^{iC}
\]
2. **Using Euler's Formula:**
By applying Euler's formula, we have:
\[
e^{i\theta} = \cos(\theta) + i \sin(\theta)
\]
Therefore, we can rewrite the equation as:
\[
2 (\cos B + i \sin B) = (\cos A + i \sin A) + (\cos C + i \sin C)
\]
3. **Separating Real and Imaginary Parts:**
From the above equation, we can separate the real and imaginary parts:
- Real part:
\[
2 \cos B = \cos A + \cos C
\]
- Imaginary part:
\[
2 \sin B = \sin A + \sin C
\]
4. **Dividing the Equations:**
We can divide the imaginary part by the real part to eliminate \( B \):
\[
\frac{2 \sin B}{2 \cos B} = \frac{\sin A + \sin C}{\cos A + \cos C}
\]
This simplifies to:
\[
\tan B = \frac{\sin A + \sin C}{\cos A + \cos C}
\]
5. **Using the Sine and Cosine Sum Formulas:**
We can use the identities for sine and cosine:
\[
\sin A + \sin C = 2 \sin\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right)
\]
\[
\cos A + \cos C = 2 \cos\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right)
\]
Substituting these into our equation gives:
\[
\tan B = \frac{2 \sin\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right)}{2 \cos\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right)}
\]
This simplifies to:
\[
\tan B = \tan\left(\frac{A+C}{2}\right)
\]
6. **Equating Angles:**
Since \( B \) is an angle in a triangle, we can equate:
\[
B = \frac{A+C}{2}
\]
Given that \( A + B + C = 180^\circ \), substituting \( B \) gives:
\[
A + \frac{A+C}{2} + C = 180^\circ
\]
Simplifying this leads to:
\[
2A + 2C = 360^\circ \Rightarrow A + C = 180^\circ
\]
This means \( A = C \).
7. **Conclusion:**
Since \( A = C \) and \( B = 60^\circ \), we conclude that all angles \( A, B, C \) are equal to \( 60^\circ \). Thus, the triangle is equilateral.
### Final Answer:
The triangle is an **equilateral triangle**.
