How to construct a 45 degree angle given its opposite segment and a line through its altitude? 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.
 A: 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 $m\angle ACB =\frac 12 m\angle ADB = 45^{\circ}$.

A: Firstly get a right angle, say $\angle ABC$. Now draw a circle with its center at $B$ with whatever radius you like. The arc will intersect $BA$ and $BC$. Say the intersection point on $BA$ is $D$ and on $BC$ is $E$. Now select a radius $r$ that is at least half the length of $DE$, draw a circle with its center at $D$ and another at $E$, and make their radius both $r$. These two circle will intersect at a point, let it be $F$. Now connect $BF$ and $\angle ABF = \angle FBC=45^{o}$.
Note 1: Why the $r$ needs to be at least $\frac 1 2 DE$? Because otherwise the two circles won't intersect.
Note 2: In fact the two circles in the last part will intersect at two points (unless $r=\frac 1 2 DE$, which will result in one intersection point). However just randomly select one and it will be fine. It can be proven that those two points and $B$ are on a straight line.
Note 3: The idea behind this procedure is to construct isosceles triangles.
A: Just bisect $90^{\circ}$ using R&C. 
To do so successfully increase compass radius/separation  to about double the first radius to be sure to get the intersection of circular arcs pair cross which to connect to right angle vertex corner at  $45^{\circ}$ to one of two arms.
