sqing wrote: The quadrilateral $ABCD$ satisfies $\angle ACD = 2\angle CAB, \angle ACB = 2\angle CAD $ and $CB = CD.$ Show that $$\angle CAB=\angle CAD.$$ Let $X = AB \cap CD$ and $Y = AD \cap BC$. Note that $\angle CAX = \angle AXC$. So, $AC = CX$. Similarly, $AC = CY$. So, $C$ is the centre of $AXY$. Note that $\triangle BCX \cong \triangle CDY$. Thus, $\angle CAB = \angle CAD$. $\blacksquare$

This is almost exactly Canada 2000/4.