Czech and Slovak Mathematical Olympiad

Assume first that $|\angle ABC| + |\angle ACM| = 90^\circ$. Using the notation $\phi = |\angle ACM|$ and $\psi = |\angle BCM|$ (Fig. 1), we conclude from our assumption that $|\angle ABC| = 90^\circ - \phi$, and hence $|\angle BAC| = 90^\circ - \psi$ as well, because of an easy angle computation in $\triangle ABC$:
$$\begin{aligned} |\angle BAC| &= 180^\circ - |\angle ABC| - |\angle ACB| \\ &= 180^\circ - (90^\circ - \phi) - (\phi + \psi) = 90^\circ - \psi. \end{aligned}$$

Fig. 1

Applying Law of Sines to $\triangle ACM$ and $\triangle BCM$, we get
$$\frac{\sin(90^\circ - \psi)}{\sin \phi} = \frac{|CM|}{|AM|} = \frac{|CM|}{|BM|} = \frac{\sin(90^\circ - \phi)}{\sin \psi}.$$
Comparing the two ratios of sines and using the formula $\sin(90^\circ - \omega) = \cos \omega$, we obtain an equality $\sin \phi \cos \phi = \sin \psi \cos \psi$ or $\sin 2\phi = \sin 2\psi$. Since the angles $\phi$ and $\psi$ are acute, both $2\phi$ and $2\psi$ are between $0^\circ$ and $180^\circ$. Thus by a well known sine property, the equality $\sin 2\phi = \sin 2\psi$ means that either $2\phi = 2\psi$ or $2\phi + 2\psi = 180^\circ$. In the first case (when $\phi = \psi$), the interior angle of $\triangle ABC$ at the vertices $A$ and $B$ are equal, in the second case (when $\phi + \psi = 90^\circ$) the interior angle at the vertex $C$ is right. This completes the proof of one of the two implications stated in the problem.

Fig. 1

To prove the second (converse) implication, let us assume that (i) $|AC| = |BC|$ or (ii) $|\angle ACB| = 90^\circ$.

Case (i). It follows from $|AC| = |BC|$ that the triangles $ACM$ and $BCM$ are congruent (by SSS theorem), with right interior angles at the vertex $M$. Consequently,

Applying Law of Sines to $\triangle ACM$ and $\triangle BCM$, we get
$$\frac{\sin(90^\circ - \psi)}{\sin \phi} = \frac{|CM|}{|AM|} = \frac{|CM|}{|BM|} = \frac{\sin(90^\circ - \phi)}{\sin \psi}.$$
The converse implication is proven.

Let $k$ be the circumcircle of the given triangle $ABC$. Its median $CM$ can be extended to the chord $CC'$ of the circle $k$ (Fig. 2). Since the inscribed angles $ABC'$ and $ACC'$ (or $ACM$) are congruent, the considered sum of angles $ABC$ and $ACM$ is equal to the angle $CBC'$. By Thales' theorem the last angle $CBC'$ is right if and only if the chord $CC'$ is a diameter of the circle $k$. This happens if and only if the centre $S$ of $k$ lies on the ray $CM$. For such a situation, we distinguish two cases: $S = M$ and $S \neq M$. Note that $S = M$ holds if and only if the angle $ACB$ is right (by Thales' theorem again). Thus let us analyse the second case $S \neq M$: The three distinct points $C, M$ and $S$ are obviously collinear if and only if the line $MS$, a perpendicular bisector of a segment $AB$, passes through the point $C$. However, the last condition is equivalent to the desired equality $|AC| = |BC|$. This completes the proof (common for the both composing implications).

Fig. 2

Remark. Instead of the chord $CC'$ of the circumcircle $k$, it is possible to consider the tangent line $t$ to the circle $k$ at its point $C$ (Fig. 3). Since the inscribed angle $ABC$ is always congruent to the marked angle between $AC$ and $t$, the sum of the angles $ABC$ and $ACM$ equals $90^\circ$ if and only if the tangent $t$ is perpendicular to the ray $CM$. The last is equivalent to the condition from the above solution, namely that the ray $CM$ passes through the centre $S$ of $k$.