Construct a triangle given a height, his base and the opposite angle

Question

Let $\overline{AB}$ and $\overline{CD}$ be given segments and $\alpha$ a given angle.

Construct the triangle $ABC$ with height $\overline{CD}$ corresponding to the side $\overline{AB}$, such that $\angle BCA=\alpha$.

Thanks in advance!

Answer 1

Find the middle of $AB$, let's call this $M$, and draw the perpendicular to $AB$ through $M$. From angle $\alpha$ we can find the angle $90^\circ-\alpha$. From $AB$ construct this angle, and call $O$ the intersection with the perpendicular through $M$. The angle $\angle AOM=\alpha$, and then the angle $\angle AOB=2\alpha$. Draw a circle with the center at $O$, and radius $OA$. All the points $N$ on the circle, on the $O$ side of the $AB$ line have $\angle ANB=\alpha$. I used the fact that the inscribed angle is half the arc, and that the arc equal to the central angle. Now on the $MO$ line draw a point $P$ at a distance $DC$ from $M$, on the same side as $O$, then draw a parallel to $AB$ through $P$. From all points on this parallel line, the perpendicular to $AB$ had the length $DC$. Now the intersections of the line with the circle obey both conditions in the problem.