, 2014

Solution 1. Since $\angle BAD = \angle DAM$ and $\angle ADB = \angle ADM = 90^\circ$, the triangles $ADB$ and $ADM$ which share the side $AD$, are congruent. Therefore $|BD| = |DM|$, and so $2|MD| = |BM| = |MC|$.
Since $\angle DAM = \angle MAC$, $AM$ bisects $\angle DAC$ and we have
$$\cos(\angle DAC) = \frac{AD}{AC} = \frac{MD}{MC} = \frac{1}{2},$$
and so $\angle DAC = 60^\circ$. It follows that $\angle ACD = 30^\circ$ and $\angle DAM = 30^\circ$.
Therefore, $\angle BAC = 90^\circ$ and $\angle ABC = 60^\circ$, i.e. the angles of the triangle
$ABC$ are $90^\circ$, $60^\circ$ and $30^\circ$.

Let $\alpha = \angle BAD = \angle DAM = \angle MAC$ and $t = \tan(\alpha)$. Then, $\tan(2\alpha) = \frac{2t}{1-t^2}$. Letting $AD = h$, we have $t = x/h$ and
$$\tan(2\alpha) = \frac{3x}{h} = \frac{2\left(\frac{x}{h}\right)}{1 - \left(\frac{x}{h}\right)^2} = \frac{2xh}{h^2 - x^2}$$
Rearranging gives $\tan(\alpha) = \frac{x}{h} = \frac{1}{\sqrt{3}}$, and so $\alpha = 30^\circ$. This implies $\angle DAC = 2\alpha = 60^\circ$, hence $\angle ACD = 30^\circ$. Finally, $\angle BAC = 3\alpha = 90^\circ$ which implies $\angle ABC = 60^\circ$.

Solution 3. Let $\alpha = \angle BAD = \angle DAM = \angle MAC$, $\beta = \angle ABM$ and $\gamma = \angle BCA$. Since $\angle BAD = \angle DAM$ and $\angle ADB = \angle ADM = 90^\circ$ we have $\angle ABM = \angle AMB$, hence $\beta = \alpha + \gamma$. Using $|BM| = |MC|$ and the Sine Theorem twice we obtain
$$\frac{\sin(\beta)}{\sin(2\alpha)} = \frac{|AM|}{|BM|} = \frac{|AM|}{|CM|} = \frac{\sin(\gamma)}{\sin(\alpha)}$$
and so
$$2 \cos(\alpha) = \frac{\sin(2\alpha)}{\sin(\alpha)} = \frac{\sin(\beta)}{\sin(\gamma)} = \frac{\sin(\alpha + \gamma)}{\sin(\gamma)}.$$
Using $\sin(\alpha + \gamma) = \sin(\alpha)\cos(\gamma) + \cos(\alpha)\sin(\gamma)$ the above simplifies to $2\cos(\alpha) = \sin(\alpha)\cot(\gamma) + \cos(\alpha)$, or equivalently $\cot(\alpha) = \cot(\gamma)$. As $\alpha$ and $\gamma$ are both less than 180°, this implies $\alpha = \gamma$. Therefore, $\beta = \alpha + \gamma = 2\alpha$ and from the right angled triangle ABD we see that $\alpha = 30^\circ$ and $\beta = 60^\circ$, hence $\gamma = 30^\circ$ and $\angle BAC = 3\alpha = 90^\circ$.