On the diagonal $AC$ of a convex quadrilateral $ABCD$ is chosen such a point $K$ such that $KD = DC$, $\angle BAC = \frac12 \angle KDC$, $\angle DAC = \frac12 \angle KBC$. Prove that $\angle KDA = \angle BCA$ or $\angle KDA = \angle KBA$.
Source: - All-Russian MO 2003 Regional (R4) 11.2
Tags: geometry, angles, equal angles
On the diagonal $AC$ of a convex quadrilateral $ABCD$ is chosen such a point $K$ such that $KD = DC$, $\angle BAC = \frac12 \angle KDC$, $\angle DAC = \frac12 \angle KBC$. Prove that $\angle KDA = \angle BCA$ or $\angle KDA = \angle KBA$.