In a convex quadrilateral $ABCD$ angles at $A,B,C$ are equal. Prove that vertex $D$ lies on the Euler line of triangle $ABC$.
My try: We can use complex numbers. Set circumcirle of triangle $ABC$ as the unit circle. Now we need to prove that $d = k(a + b + c + d)$. I just don't know how to handle the angle condition (I have thought of connecting $D$ with $A$ and $C$ via complex spiral similarity but that's already lot of work and I think this kind of problem should have a much simpler solution, I guess I just don't like these computational techniques :) ).
Let $AD$ and $BC$ meets at $M$ and let $AB$ and $CD$ meet at $N$. Let $A'$ be a midpoint of $BC$ and $C'$ be a midpoint of $AB$.
So $MC'$ and $A'N$ are perpendicular to $AB$ and $BC$, so they meet at $S$, the circumcenter of $ABC$.
Lines $AA'$ and $CC'$ meet at $G$ the gravity center for $ABC$.
Now by Pappus theorem $S,G$ and $D$ are collinear.