, 2010

Denote $\angle ADB = \angle DCA = \alpha$ and $\angle CBD = \angle BAC = \beta$ (Fig. 1).

In triangle $DAC$ we have $|DA| = |DC|$ and therefore $\angle DAC = \angle DCA = \alpha$; analogously in triangles $DAB$ and $DBC$, we have $\angle DBA = \angle DAB = \alpha + \beta$ and $\angle DCB = \angle DBC = \beta$, respectively. So $\angle BCA = \beta - \alpha$.

From triangle $ABC$ now $\beta + \alpha + \beta + \beta + \beta - \alpha = 180^\circ$ or, equivalently, $4\beta = 180^\circ$, giving $\beta = 45^\circ$.

From triangle $ADB$ we get $\alpha + \beta + \alpha + \beta + \alpha = 180^\circ$ or, equivalently, $3\alpha = 180^\circ - 2\beta = 90^\circ$ and $\alpha = 30^\circ$.

Therefore, the sizes of the angles of quadrangle $ABCD$ are $\angle DAB = \alpha + \beta = 75^\circ$, $\angle ABC = \alpha + 2\beta = 120^\circ$, $\angle BCD = \beta = 45^\circ$, and $\angle CDA = 360^\circ - 75^\circ - 120^\circ - 45^\circ = 120^\circ$.

Fig. 1

We use the same notation as in the Solution 1. Triangle $BCD$ is isosceles, hence $\angle DCB = \angle DBC = \beta$. As $D$ is the circumcenter of $ABC$, we have $\angle BDC = 2\angle BAC = 2\beta$. The sizes of the angles of triangle $BCD$ are therefore $\beta, \beta$, and $2\beta$; thus $\beta + \beta + 2\beta = 180^\circ$, whence $\beta = 45^\circ$.

As $\angle BCA = \frac{\angle BDA}{2}$, we have $\angle BCD = \frac{\alpha}{2} + \alpha = \beta$, whence $\alpha = \frac{2}{3}\beta = 30^\circ$.

Consequently, the sizes of the angles of quadrangle $ABCD$ are $\angle DAB = \angle ABD = \frac{180^\circ - \alpha}{2} = 75^\circ$, $\angle ABC = \angle ABD + \angle CBD = 75^\circ + 45^\circ = 120^\circ$, $\angle BCD = \beta = 45^\circ$, and $\angle CDA = \angle CDB + \angle BDA = 90^\circ + 30^\circ = 120^\circ$.