19. Construct a triangle given the median, altitude, and angle bisector drawn from the same vertex.
19. Let $m_{a}, \beta_{a}$, and $h_{a}$ be the median, bisector, and altitude of the desired triangle $ABC$, drawn from vertex $A$. We construct a right triangle $ADC$ with hypotenuse $AE = \beta_{a}$ and leg $AD = h_{a}$, with a right angle at vertex $D$. Next, from vertex $A$ as the center and with radius equal to $m_{a}$, we draw a circle that intersects ray $DE$ at point $F$. We find the point $K$ of intersection of line $AE$ with the perpendicular $FK$ to line $DE$ at point $F$. Through the midpoint of segment $AK$, we draw a perpendicular until it intersects line $FK$ at point $O$. This point will be the center of the circumscribed circle around the desired triangle $ABC$. The points of intersection of this circle with line $DE$ will be the vertices $B$ and $C$ of triangle $ABC$. The necessary condition for the solvability of the problem is the relationship $m_{a} \geqslant \beta_{a} \geqslant h_{a}$.

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