I need to do this construction for perspective geometry and art. So I'm only using a straight edge and compass.
I can solve the problem if the angle needs to be 90 degrees. I just draw a semicircle with the segment as diameter. Then, where the semicircle intersects the perpendicular, I draw a triangle between this new vertex and the segment. This gives the required triangle.
The analogous construction for a 45 degree angle seems harder. Is there a better way?
Edit:
I think I didn't explain well.
I want to take this data
And construct the following:
Given an altitude, and a segment with vertices A and B, I want to find a point C on the altitude such that $\angle ACB$ is 45 degrees. It's not clear to me how to find a 90 degree angle, such that after bisection, both of the rays of a 45 intersect A and B.
Best Answer
Use this fact:
If $M,N$ are point's on a circle. And $O$ is the center of the circle. Then if $P$ is a point on the circle that on the same side of $\overline MN$ as $O$ the center then $m\angle MPN = \frac 12 m\angle MON$.
So construct the perpendicular bisector of $\overline {AB}$ (which will be parallel to but will not be the altitude). Find the point $D$ on the perpendicular bisector so that $m\angle ADB = 90^{\circ}$. Now $AD=DB$ as $D$ is on the perpendicular bisector of $\overline {AB}$ so $A$ and $B$ are points of a circle with radius $AD=DB$ with center $D$.
Find the point $C$ on the altitude so that $DC = AD=AB$. (There will be two such points, probably, but one will be below $\overline {AB}$ and the other above-- choose the one that is above.)
Then $A,C,B$ are three points on the circle so $m\angle ACB =\frac 12 m\angle ADB = 45^{\circ}$.