Let $A_1A_2A_3\cdots A_{10}$ be a regular decagon and $A=A_1A_4\cap A_2A_5, B=A_1A_6\cap A_2A_7, C=A_1A_9\cap A_2A_{10}.$ Find the angles of the triangle $ABC$.
1. **Representation in the Argand Plane:**
   We start by representing the vertices of the regular decagon in the complex plane. Let \( z = \cos\left(\frac{\pi}{5}\right) + i\sin\left(\frac{\pi}{5}\right) \). The vertices \( A_i \) of the decagon can be represented as \( A_i = z^{i-1} \) for \( i \in \{1, 2, \ldots, 10\} \).

2. **Finding Intersection Points:**
   - For point \( A \):
     \[
     A = A_1A_4 \cap A_2A_5
     \]
     Using the coordinates of \( A_1, A_4, A_2, \) and \( A_5 \):
     \[
     A_1 = 1, \quad A_4 = z^3, \quad A_2 = z, \quad A_5 = z^4
     \]
     The intersection point \( A \) can be found using the formula for the intersection of two lines in the complex plane:
     \[
     A = \frac{z^3(z + z^4) - z^5(1 + z^3)}{z^3 - z^5} = \frac{z(z^3 + 1)}{z + 1}
     \]

   - For point \( B \):
     \[
     B = A_1A_6 \cap A_2A_7
     \]
     Using the coordinates of \( A_1, A_6, A_2, \) and \( A_7 \):
     \[
     A_1 = 1, \quad A_6 = z^5, \quad A_2 = z, \quad A_7 = z^6
     \]
     The intersection point \( B \) is:
     \[
     B = 0
     \]

   - For point \( C \):
     \[
     C = A_1A_9 \cap A_2A_{10}
     \]
     Using the coordinates of \( A_1, A_9, A_2, \) and \( A_{10} \):
     \[
     A_1 = 1, \quad A_9 = z^8, \quad A_2 = z, \quad A_{10} = z^9
     \]
     The intersection point \( C \) is:
     \[
     C = \frac{z^8(z + z^9) - z^{10}(1 + z^8)}{z^8 - z^{10}} = \frac{z(z^8 + 1)}{z + 1}
     \]

3. **Simplifying the Points:**
   Let us simplify the points \( A, B, \) and \( C \):
   - \( A = \frac{z(z^3 + 1)}{z + 1} \)
   - \( B = 0 \)
   - \( C = \frac{z(z^8 + 1)}{z + 1} \)

4. **Transforming to Simpler Triangle:**
   Let us consider a simpler triangle \( \triangle A'B'C' \) with:
   - \( A' = 0 \)
   - \( B' = \frac{1 - z^2}{2} \)
   - \( C' = 1 \)

   We can see that \( \triangle ABC \sim \triangle A'B'C' \).

5. **Calculating Angles:**
   - \( 1 - z^2 = 1 - \cos\left(\frac{2\pi}{5}\right) - i\sin\left(\frac{2\pi}{5}\right) = -2i\sin\left(\frac{\pi}{5}\right)z \)
   - \( 1 + z^2 = 1 + \cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right)z \)

   - \( \angle B'A'C' = \arg(-z) + \arg\left(i\sin\left(\frac{\pi}{5}\right)\right) = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10} \)
   - \( \angle A'B'C' = \arg\left(\frac{z^2 - 1}{z^2 + 1}\right) = \arg\left(i\tan\left(\frac{\pi}{5}\right)\right) = \frac{\pi}{2} \)
   - \( \angle A'C'B' = \frac{\pi}{5} \)

Thus, the angles of \( \triangle ABC \) are \( \frac{3\pi}{10}, \frac{\pi}{2}, \frac{\pi}{5} \).

The final answer is \( \boxed{ \frac{3\pi}{10}, \frac{\pi}{2}, \frac{\pi}{5} } \)