10.91*. In an acute-angled triangle $A B C$, the bisector $A D$, the median $B M$, and the altitude $C H$ intersect at one point. Within what limits can the measure of angle $A$ vary?
10.91. Draw a perpendicular to side $AB$ through point $B$. Let $F$ be the point of intersection of this perpendicular with the extension of side $AC$ (Fig. 10.3). We will prove that the bisector $AD$, the median $BM$, and the altitude $CH$ intersect at one point if and only if $AB = CF$. Indeed, let $L$ be the point of intersection of $BM$ and $CH$. The bisector $AD$ passes through point $L$ if and only if $BA: AM = BL: LM$, but $BL: LM = FC: CM = FC: AM$.

![](https://cdn.mathpix.com/cropped/2024_05_21_22b0ba5e8c5a1f49bd81g-254.jpg?height=366&width=783&top_left_y=237&top_left_x=755)

Fig. 10.3

If on side $AF$ of some right triangle $ABF$ ($\angle ABF = 90^\circ$) we lay off the segment $CF = AB$, then the angles $BAC$ and $ABC$ will be acute. It remains to determine in which cases the angle $ACB$ will be acute. Drop a perpendicular $BP$ from point $B$ to side $AF$. The angle $ACB$ is acute if $FP > FC = AB$, i.e., $BF \sin A > BF \cot A$. Therefore, $1 - \cos^2 A = \sin^2 A > \cos A$, i.e., $\cos A < \arccos((\sqrt{5-1}) / 2) \approx 51^\circ 50'$.