, 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 $\triangle ADC$ is an isosceles triangle and $AD = CD = a$. Then $\triangle AOD$ 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 $\triangle BOC$ is an isosceles triangle, so $OC = BC = b$. Thus, $\triangle COD$ is also an isosceles triangle ($OC = b$ and $OD = b$). Therefore, $\angle ODC = \angle OCD = \alpha$.

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

In $\triangle BCD$ 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 $\triangle ADB$ 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, $\angle B = 180^\circ - \angle A = 180^\circ - 54^\circ = 126^\circ$.