Problem 9. In a convex quadrilateral $A B C D$, side $A B$ is equal to diagonal $B D, \angle A=65^{\circ}$, $\angle B=80^{\circ}, \angle C=75^{\circ}$. What is $\angle C A D$ (in degrees $) ?$
Answer: 15.

Solution. Since triangle $ABD$ is isosceles, then $\angle BDA = \angle BAD = 65^{\circ}$. Therefore, $\angle DBA = 180^{\circ} - 130^{\circ} = 50^{\circ}$. Hence, $\angle CBD = 80^{\circ} - 50^{\circ} = 30^{\circ}$, $\angle CDB = 180^{\circ} - 75^{\circ} - 30^{\circ} = 75^{\circ}$. This means that triangle $DBC$ is isosceles, so $BC = BD = AB$. Therefore, $\angle BCA = \angle BAC = \frac{1}{2}(180^{\circ} - 80^{\circ}) = 50^{\circ}$. Hence, $\angle CAD = 65^{\circ} - 50^{\circ} = 15^{\circ}$.