Construct a triangle, given the angle at the vertex, the altitude, and the ratio in which its foot divides the base. The problem is from Kiselev's Geometry exercise 392:

Construct a triangle, given the angle at the vertex, the altitude, and
the ratio in which its foot divides the base.

The chapter is about homothety. My attempt was to first omit the altitude condition and start from an arbitrary segment that is divided by the ratio. From the division point, erect a perpendicular line. The problem will be solved by using homothety if an angle congruent to the given angle is constructed whose vertex lies on the perpendicular line and which passes through the endpoints of the segment. Unfortunately, I could not achieve this.
Any help would be much appreciated.
 A: I assume that you want a purely geometrical construction (not analytical). So you are given an angle $\alpha$. Let's call the vertex $O$.

*

*You can easily create an angle $2\alpha$, with the vertex at $O$.

*Draw a circle with the center at $O$, of any radius, say $R$, and call $A$ and $B$ the intersections of the circle with the sides of the angle $2\alpha$. Now any point on the circle on the right side of $AB$ will see the segment at an angle $\alpha$ (on the other side it will be $180^\circ-\alpha$).

*Divide the segment $AB$ in the correct ratio. I assume that you have a different segment already divided in this ratio, let's call it $A'B'$, with $C'$ in between $A'$ and $B'$, so it is divided in the correct ratio. Move $A'$ to $A$, such as that $B'$ is not along $AB$. Draw the $BB'$ line, and a line through $C'$ parallel to it. It will intersect the $AB$ segment at $C$, which will divide the $AB$ in the desired ratio.

*Draw the perpendicular to $AB$ at $C$, and the intersection with the circle will be point $D$. You have $\angle ADB=\alpha$, $AC/CB$is the correct ratio, and $DC\perp AB$.

*The last step is to take a segment along $DC$, away from $D$ of the desired length for the altitude, call $P$ the point at this distance. Draw a parallel to $AB$ through $P$. The intersections of this line with $AD$ and $BD$ will give you the other two vertices of your triangle.


