In the quadrilateral $ABCD$ it is known that $ABC + DBC = 180^o$ and $ADC + BDC = 180^o$. Prove that the center of the circle circumscribed around the triangle $BCD$ lies on the diagonal $AC$.
Source:
Tags: angles, Circumcenter, geometry
In the quadrilateral $ABCD$ it is known that $ABC + DBC = 180^o$ and $ADC + BDC = 180^o$. Prove that the center of the circle circumscribed around the triangle $BCD$ lies on the diagonal $AC$.