Construction with compass and straight edge of a triangle.

How can I construct a triangle given the altitudes of side $a$ and side $b$ and the angle $\beta$ at vertex $B$?

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Let $p$ be the given altitude to the side $a$, and let $q$ be the given altitude to the side $b$.

We produce an algebraically motivated solution. The idea is that we first produce a triangle similar to the desired triangle, and then scale it to the desired size.

Draw an arbitrary line segment, which we call $A'B'$. At $B'$, draw a line $\ell$ that makes angle $\beta$ with line $A'B'$. Drop a perpendicular from $A'$ to the line $\ell$, meeting $\ell$ at say $M'$

Let $p'$ be the length of the just drawn perpendicular. Construct a line segment of length $q'=\frac{p'}{p}q$. This can be done by compass and straightedge.

Now draw the circle $X$ with centre the midpoint of $A'B'$, and diameter $A'B'$. Draw the circle with centre $B'$ and radius $q'$. These circles meet at a point $N'$ such that the altitude from $B'$ to $A'N'$ has length $q'$. Draw the line $A'N'$, and let it meet $\ell=B'M'$ at $C'$.

The triangle $A'B'C'$ has the shape that we want, except for being wrong by a scaling factor of $p'/p$. Scale it up, by multiplying the sides by $p/p'$. This can be done by straightedge and compass.

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You could just hold off on defining $A'$ until after you've drawn $\ell$ at angle $\beta$. Draw a line parallel to $\ell$ at a distance $p$, and the point where it meets the first line through $B'$ is the point $A'$ with perpendicular $p$ by construction. Then you don't have to rescale the triangle in the end. – Rahul Nov 12 '12 at 7:56
@RahulNarain: Yes, good observation, which will be helpful to the OP. You may wish to write up a separate solution. I will keep mine as is, because I want to keep the idea of scaling. – André Nicolas Nov 12 '12 at 8:01
Thank you for your reply! – HowardRoark Nov 13 '12 at 5:41

Let the lengths of the altitudes to the sides $a$ and $b$ be $p$ and $q$.

1. Draw a line $\ell_C$. Pick a point $B$ on it, and draw line $\ell_A$ through $B$ at angle $\beta$ to $\ell_C$.

Lines $\ell_A$ and $\ell_C$ form two of the sides of the triangle.

2. Draw a line parallel to $\ell_A$ at distance $p$, meeting $\ell_C$ at point $A$.

Now the altitude from $A$ to $\ell_A$ has length $p$.

3. Draw a circle of radius $q$ centered at $B$. Draw line $\ell_B$ through $A$ tangent to this circle, meeting $\ell_A$ at point $C$.

Now the altitude from $B$ to $\ell_B$ has length $q$.

Thus, $\triangle ABC$ is the desired triangle. (This is essentially the same as André Nicolas's construction with the modification I suggested in a comment.)

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Thank you! I did end up getting the solution after all! I actually did exactly what you outlined here! Awesome! – HowardRoark Nov 13 '12 at 5:41