National Math Olympiad – Final Round, 2015

Solution 1. Notice that the triangle $ABC$ is isosceles with apex at vertex $B$. Let $E$ be such a point on the line $CD$ that $\angle CAE = 30^\circ$, so that the triangle $ACE$ is also isosceles with apex at vertex $E$. Since $\angle CAD = 40^\circ$ the point $E$ lies between $C$ and $D$. Thus the quadrilateral $ABCE$ is a deltoid and its diagonals $AC$ and $BE$ intersect at a right-angle. Therefore we have $\angle BEC = 90^\circ - 30^\circ = 60^\circ$ and hence $\angle BAD = 60^\circ = \angle BEC = 180^\circ - \angle DEB$. This means that the points $A, B, E$, and $D$ are concyclic. By Angles Subtended by Same Arc Theorem we have $\angle EBD = \angle EAD = 40^\circ - 30^\circ = 10^\circ$, hence $\angle CBD = \angle CBE + \angle EBD = (90^\circ - 20^\circ) + 10^\circ = 80^\circ$.

Solution 2. Notice that the triangle $ABC$ is isosceles with apex at vertex $B$. Let $\mathcal{K}$ be the circle with center $B$ which passes through the points $A$ and $C$. Let's prove that the point $D$ also lies on the circle $\mathcal{K}$. From the data in the problem we calculate $\angle ADC = 180^\circ - 40^\circ - 30^\circ = 110^\circ$ and $\angle CBA = 180^\circ - 20^\circ - 20^\circ = 140^\circ$. Therefore the central angle subtended by the arc $\widehat{AC}$ equals $\angle ABC = 360^\circ - \angle CBA = 220^\circ$ and is double the angle $\angle ADC$. By the Angle at the Center Theorem the angle $\angle ADC$ is the inscribed angle subtended by the arc $\widehat{AC}$, which means that the point $D$ lies on the circle $\mathcal{K}$. From this it follows that the triangle $CBD$ is isosceles with apex at vertex $B$ and hence $\angle CBD = 180^\circ - 2\angle DCB = 180^\circ - 100^\circ = 80^\circ$.