On the side $AB$ of the equilateral triangle $ABC$, $8$ points $D_1, D_2, ..., D_8$ are chosen that divide the side $AB$ into equal parts, that is, $AD_1=D_1D_2 = ... D_8B$. On the $BC$ side, we chose a point $P$ such that $AD_1 = CP$. Find the sum of the angles $$\angle CD_1P + \angle CD_2P + ... + \angle CD_8P$$
My work so far:
$$AC||D_1P \Rightarrow \angle CD_1P =\angle ACD_1$$
I thought then $$\angle CD_1P + \angle CD_2P + ... + \angle CD_8P=\angle ACD_1+\angle D_1CD_2+\angle D_2CD_3+...=\angle ACD_8,$$ but $$CD_1 \not||PD_2$$ then $$\angle CD_2P\not=\angle D_1CD_2$$
Notice we can rewrite the sum of angles as $$\begin{align}\sum_{k=1}^8 \angle CD_kP &= \sum_{k=1}^8 (\angle CD_kB - \angle PD_kB) = \sum_{k=1}^8 \angle CD_kB - \sum_{k=1}^8 \angle PD_kB\\ &= \underbrace{\sum_{k=1}^8 \angle CD_kB}_{\mathcal{A}} - \angle PD_1B - \underbrace{\sum_{k=2}^8 \angle PD_kB}_{\mathcal{B}} \end{align} $$
Since $\angle CD_kB + \angle CD_{9-k}B = 180^\circ$ for $k = 1, .., 8$, we have
$$\mathcal{A} = \sum_{k=1}^8\angle CD_k B = \frac12\sum_{k=1}^8(\angle CD_k B + \angle CD_{9-k}B) = 8 \times 90^\circ$$
Since $\triangle PD_1B$ is equilateral, we have $\angle PD_1 B = 60^\circ$.
Since $\angle PD_kB + \angle PD_{10-k}B = 180^\circ$ for $k = 2,..8$, we have $$\mathcal{B} = \sum_{k=2}^8 \angle PD_kB = \frac12\sum_{k=2}^8 (\angle PD_kB + \angle PD_{10-k}B) = 7 \times 90^\circ$$
Combine these, we obtain:
$$\sum_{k=1}^8 \angle CD_kP = 8 \times 90^\circ - 60^\circ - 7 \times 90^\circ = 30^\circ $$