Given is acute triangle $ABC$. Let $D$ be foot of altitude from vertex $A$. Let $D_1$ be a point so that line of symmetry between $D_1$ and $D$ is line $AB$. Let $D_2$ be a point so that line of symmetry between $D_2$ and $D$ is line $AC$. Let points $E_1, E_2$ be on line $BC$ so that $D_1E_1 \parallel AB$ and $D_2E_2 \parallel AC$. Proof that points $D_1, E_1, D_2, E_2$ lie on same circle and that center of this circle lies on the circumscribed circle of triangle $ABC$.
My plan was to first prove that $D_1E_1E_2D_2$ is cyclic quadrilateral, in other words that $\angle E_1E_2D_2 + \angle E_1D_1D_2 = 180°$ or that $\angle E_1D_2D_1 = \angle E_1E_2D_1$. This would mean that points $D_1, E_1, D_2, E_2$ lie on same circle. However, without success.
Can someone help me with this please?