Belorusija 2012

Answer: $\angle A=54^{\circ }$, $\angle B=126^{\circ }$, $\angle C=72^{\circ }$, $\angle D=108^{\circ }$. Let $CD=AO=a$, $BC=OD=b$ and $\angle BCD=2\alpha$. By condition,



$BD=y$, $\angle DCA=\angle ACB=\alpha$. Since $BC\parallel AD$, we have $\angle CAD=\angle ACB=\alpha$. So $△ADC$ is an isosceles triangle and $AD=CD=a$. Then $△AOD$ is also an isosceles triangle ($AO=a$ and $AD=a$). Therefore, $\angle AOD=\angle ADO$.

Further, since $\angle BOC=\angle AOD$ (vertical angles) and $\angle OBC=\angle ODA$ (alternate angles for $BC\parallel AD$), we see that $△BOC$ is an isosceles triangle, so $OC=BC=b$. Thus, $△COD$ is also an isosceles triangle ($OC=b$ and $OD=b$). Therefore, $\angle ODC=\angle OCD=\alpha$.

Then in $△COD$ the external angle $\angle BOC=\angle OCD+\angle ODC=\alpha +\alpha =2\alpha$, hence, $\angle OBC=\angle BOC=2\alpha$. It means that $△BCD$ is an isosceles triangle ($\angle DBC=2\alpha$ and $\angle BC=2\alpha$). Then $BD=CD=a=AD$ and so $△ADB$ is also an isosceles triangle.

In $△BCD$ the sum $\angle BCD+\angle CDB+\angle DBC=2\alpha +\alpha +2\alpha =5\alpha$. Since the sum of all angles of a triangle is equal to $180^{\circ }$, we have $5\alpha =180^{\circ }$, hence $\alpha =36^{\circ }$.

Then for the trapezoid $ABCD$ we have $\angle C=2\alpha =72^{\circ }$, $\angle D=180^{\circ }-\angle C=180^{\circ }-72^{\circ }=108^{\circ }$. Since $△ADB$ is isosceles, we have $\angle A=\frac{1}{2}(180^{\circ }-\angle ADB)=\frac{1}{2}(180^{\circ }-2\alpha )=90^{\circ }-\alpha =90^{\circ }-36^{\circ }=54^{\circ }$. Hence, $\angle B=180^{\circ }-\angle A=180^{\circ }-54^{\circ }=126^{\circ }$.