Minimal-area rectangle, bounding (covering) an acute triangle Given acute triangle $ABC$ such that $AB$ is the longest side. I need to prove that the rectangle $PQRS$ of minimum area covering the whole of triangle $ABC$
has one side (say $PQ$) coincident with $AB$ and the opposite side $RS$ parallel to $PQ$ and passing through $C$.
(The application I have that needs this proof does not guarantee that $ABC$ is an acute triangle, only that point $C$ lies inside the strip defined by  the two perpendiculars to $AB$ passing through points $A$ and $B$.  But if angle $C$ is obtuse, it is much easier to prove that the minimal bounding rectangle is as stated.)
If the statement is untrue, I'd like to know the conditions for it to be wrong.
This is not as trivial (for a generic acute triangle) as it looks.  I can show that the minimal bounding rectangle contains $A$, $B$ and $C$ on its boundary, since otherwise you could "slide" a side of the purported minimal rectangle inward until it does touch one of the vertices of the triangle and get a smaller bounding rectangle. I believe that the last such "sliding" step also shows that one of the vertices on the triangle must coincide with a vertex of the minimal bounding rectangle.
That is how far I have gotten in the forward direction.  Working backwards, I think if one shows that one side of the minimal bounding rectangle coincides with a side of the triangle, it will not be hard to show that it would have to be the longest side.
Help would be appreciated.

Note that in the case of an acute triangle you can align any side of the triangle with a side of the rectangle, and obtain the same minimal rectangle area, namely twice the area of the triangle.
I'm referring to the following figure:

We turn a circumscribing rectangle clockwise around the triangle so that when $\phi=0$ a side of the rectangle coincides with the side $AB$ of the triangle. The area $F$ of the rectangle then computes to
$$F=bc\cos\phi\cos\theta={bc\over2}\bigl(\cos(\phi+\theta)+\cos(\phi-\theta)\bigr)={bc\over2}\bigl(\sin\alpha+\cos(\phi-\theta)\bigr)\ .$$
This is obviously smallest if $|\phi-\theta|$ is largest, namely $={\displaystyle{\pi\over2}}-\alpha$. In this way we obtain $$F_{\min}=bc\sin\alpha=2{\rm area}(\triangle)\ ,$$ as claimed.
