Let us denote in a standard way the sides of triangle $a, b, c$. Then under condition, $a > b > c$. Let us denote the height and median from the top $B$ as $h_b = BH$ and $m_b = BM$ accordingly (fig. 48). Then the interesting for us angle is $\angle MBH$, while $\cos\angle MBH = \frac{h_b}{m_b}$. Similarly, we can determine cosines of the other studied angles. In order to prove that the angle at the top $B$ is

Fig. 48

the biggest, it should be enough to show that cosine of this angle is the smallest, that is:
$$\frac{h_b}{m_b} < \min \left\{ \frac{h_a}{m_a}, \frac{h_c}{m_c} \right\} . \text{ For this, let us prove that}\$$
$$\frac{m_b^2}{h_b^2} > \frac{m_a^2}{h_a^2} \quad \text{ta}\quad \frac{m_b^2}{h_b^2} > \frac{m_c^2}{h_c^2}.$$
To do this, let us use the following formula to calculate the height and median:
$$h_a = \frac{2S_{ABC}}{a} \text{ ta } m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}.$$
The first inequation is equivalent to the following:
$$\frac{2a^2 + 2c^2 - b^2}{4} \cdot \frac{b^2}{4S^2} > \frac{2b^2 + 2c^2 - a^2}{4} \cdot \frac{a^2}{4S^2} \Leftrightarrow (2a^2 + 2c^2 - b^2)b^2 > (2b^2 + 2c^2 - a^2)a^2 \Leftrightarrow 2b^2c^2 - b^4 > 2a^2c^2 - a^4 \Leftrightarrow (a^2 + b^2)(a^2 - b^2) > 2c^2(a^2 - b^2).$$
The last inequation is true, since under the condition: $a > b > c$.
Similarly, for the second ineqation we have:
$$(2a^2 + 2c^2 - b^2)b^2 > (2b^2 + 2a^2 - c^2)c^2 \Leftrightarrow 2b^2a^2 - b^4 > 2a^2c^2 - c^4 \Leftrightarrow 2a^2(b^2 - c^2) > (b^2 + c^2)(b^2 - c^2).$$
The last inequation, again, is true since $a > b > c$.