67th Romanian Mathematical Olympiad

Case 1: $D$ and $A$ are on different sides of $BC$ (figure 1). Denote $\{E\}=AC\cap DB$. Then $m(\widehat{ABE})=45^{\circ }$, therefore $[BA]$ is bisector and altitude in the triangle $BEC$. So, $[AE]=[AC]$. Construct $AM\perp BE$, Then $[AM]$ is a midline in the triangle $EBC$, hence $AM=\frac{1}{2}BC=\frac{1}{2}AD$. In the right triangle $MAD$ the leg $[AM]$ is half of the hypotenuse $[AD]$, hence $m(\widehat{ADB})=30^{\circ }$. It follows $m(\widehat{BAD})=180^{\circ }-m(\widehat{ADB})-m(\widehat{ABD})=180^{\circ }-30^{\circ }-105^{\circ }=15^{\circ }$.



Case 2: $D$ and $A$ are on different sides of the line $BC$ (figure 2). Denote $\{E\}=AC\cap DB$. Then $m(\widehat{ABE})=45^{\circ }$, hence $[BA]$ is bisector and altitude in the triangle $BEC$. So, $[AE]=[AC]$. Construct $AM\perp BE$. Then $[AM]$ is a midline in the triangle $EBC$, hence $AM=\frac{1}{2}BC=\frac{1}{2}AD$. In the right triangle $MAD$ the leg $[AM]$ is half of the hypotenuse $[AD]$, hence $m(\widehat{ADB})=30^{\circ }$. It follows $m(\widehat{BAD})=180^{\circ }-m(\widehat{ADB})-m(\widehat{ABD})=180^{\circ }-30^{\circ }-45^{\circ }=105^{\circ }$.