Germany TST

Solution:

The vertices of the rhombus are named $K\ (K \in AE)$, $L\ (L \in AB)$, $M\ (M \in BD)$ and $N\ (N \in DE)$. Let $d(X, YZ)$ denote the distance of a point $X$ from a line $YZ$. Because $D$ and $E$ lie on the respective angle bisectors, we have $d(D, AB) = d(D, AC)$, $d(E, AB) = d(E, BC)$ and $d(D, BC) = d(E, AC) = 0$, from which $d(D, AC) + d(D, BC) = d(D, AB)$ and $d(E, AC) + d(E, BC) = d(E, AB)$ follow.

Because $N$ lies on the segment $DE$ and because in the equation $d(X, AC) + d(X, BC) = d(X, AB)$, as $X$ moves along the segment $DE$, all terms change only linearly, it follows

from the two relations above also for $N$:

$d(N, AC) + d(N, BC) = d(N, AB) \qquad (1)$.

Using the notation of the figure, we have $d(N, AC) = s \cdot \sin \mu$ and $d(N, BC) = s \cdot \sin v$. Because the rhombus $KLMN$ lies entirely in one of the half-planes with respect to $AB$, we obtain from its parallelogram property

$d(N, AB) = d(N, AB) + d(L, AB) = d(K, AB) + d(M, AB) = s(\sin \delta + \sin \varepsilon)$.

With (1) it follows that $\sin \mu + \sin v = \sin \delta + \sin \varepsilon$ $\qquad$ (2).

From the assumption $\varphi > \max (\alpha, \beta)$, because of $\text{+ = CKL = +}$, it would directly follow that $\mu = \alpha - \varphi + \delta < \delta$ and analogously $v < \varepsilon$. Because $KL \parallel MN$, we have $\beta = \delta + v$, hence $\delta < \beta < 90^\circ$. Similarly it follows that $\varepsilon < 90^\circ$. Thus we obtain $\sin \mu < \sin \delta$ and $\sin v < \sin \varepsilon$, in contradiction to (2). Therefore $\varphi \leq \max (\alpha, \beta)$ holds.