[ Inequalities for elements of a triangle (other)]

In an acute-angled triangle $ABC$, the bisector $AD$, the median $BM$, and the altitude $CH$ intersect at one point. Within what limits can the measure of angle $A$ vary?
Let's draw a perpendicular from point $B$ to side $A B$. Let $F$ be the point of intersection of this perpendicular with the extension of side $A C$ (see figure). We will prove that the bisector $A D$, the median $B M$, and the altitude $C H$ intersect at one point if and only if $A B = C F$. Indeed, let $L$ be the point of intersection of $B M$ and $C H$.

The bisector $A D$ passes through point $L$ if and only if $B A : A M = B L : L M$, but $B L : L M = F C : C M = F C : A M$.

If on side $A F$ of some right triangle $A B F$ ( $\angle A B F = 90^{\circ}$ ) we lay off the segment $C F = A B$, then the angles $B A C$ and $A B C$ will be acute. It remains to determine in which cases the angle $A C B$ will be acute. Drop a perpendicular $B P$ from point $B$ to side $A F$. The angle $A C B$ is acute if $F P > F C = A B$, i.e., $B F \sin A > B F \cot A$.

Therefore, $1 - \cos^2 A = \sin^2 A > \cos A$, i.e., $\cos A < \frac{\sqrt{5} - 1}{2} \approx 0.618$, which means $\angle A > \arccos \left( \frac{\sqrt{5} - 1}{2} \right) \approx 51^\circ 50'$.

![](https://cdn.mathpix.com/cropped/2024_05_06_08342b966a7aba566077g-39.jpg?height=595&width=1309&top_left_y=2035&top_left_x=380)

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