Final Round of National Olympiad

There is clearly only one way to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $E$ and $F$ would be symmetrical with respect to the diagonal $BD$ and $\angle EDF = 60^\circ$. Thus it is possible in Malle's rhombus to choose $E$ and $F$ asymmetrically with respect to diagonal $BD$ such that triangle $DEF$ is equilateral; let us consider one of the possible setups. W.l.o.g., we can assume that $|EB| < |BF|$ (otherwise we can switch the roles of $A$ and $C$ and the roles of $E$ and $F$). Let $\alpha = \angle BAD = \angle BCD$ and $\beta = \angle AED$. Also let $E'$ be the point symmetrical to $E$ with respect to diagonal $BD$ (see fig. 17); then $|DE'| = |DE| = |DF|$ gives that triangle $E'DF$

is isosceles and $\angle BFD = \angle FE'D = \angle AED = \beta$. Therefore
$$\angle CDF = 180^\circ - \angle DCF - \angle DFC = 180^\circ - \alpha - (180^\circ - \beta) = \beta - \alpha.$$
Now
$$\begin{aligned} 180^\circ - \alpha &= \angle ADC = \angle ADE + \angle EDF + \angle FDC \\ &= (180^\circ - \alpha - \beta) + 60^\circ + (\beta - \alpha) = 240^\circ - 2\alpha. \end{aligned}$$
From here $\alpha = 60^\circ$, that is the angles of the rhombus are $60^\circ$ and $120^\circ$.

Figure 18

Let the first choices of Malle be $E_1$ and $F_1$ and second ones $E_2$ and $F_2$ (see fig. 18). As $\angle E_1DF_1 = 60^\circ = \angle E_2DF_2$ and $|DE_1| = |DF_1|$ and $|DE_2| = |DF_2|$, the rotation of the plane by $60^\circ$ around point $D$ that takes $E_1$ to $F_1$ must also take $E_2$ to $F_2$. The same rotation must then take the line $E_1E_2$ to $F_1F_2$ or line $AB$ to line $BC$. Therefore $\angle ABC = 120^\circ$, thus the angles of the rhombus are $60^\circ$ and $120^\circ$.

Figure 18