On the sides $AC$ and $AB$ of the triangle $ABC$, the points $D$ and $E$ were chosen such that $\angle ABD =\angle CBD$ and $3 \angle ACE = 2\angle BCE$. Let $H$ be the point of intersection of $BD$ and $CE$, and $CD = DE = CH$. Find the angles of triangle $ABC$.
Problem
Source: 2008 All-Ukrainian Correspondence MO of magazine ''In the World of Mathematics'', grades 5-12 p7
Tags: geometry, equal angles, equal segments, Ukraine Correspondence