The Dido problem with an arclength constraint

It is well known that the solution to the classical Dido problem is a semicircle, and that the solution to the classical isoperimetric problem is a circle. It's also reasonably obvious that the solution to the following variant is a circular arc:

Let $A$ and $B$ be fixed points on a plane, and let $l$ be a length greater than $\overline{AB}$. Which (smooth) curve through $A$ and $B$, of length $l$, maximises the area between itself and the line $AB$?

It's a straightforward exercise to verify extremality using the calculus of variations, but are there alternative proofs that do not invoke e.g. the Euler-Lagrange equations? This was originally a homework problem with the isoperimetric inequality given as a hint, and I'm just wondering what the intended solution was...

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For any length $l > \overline{AB}$, we can find a circle passing through $AB$ such that the length of a circular arc between $A$ and $B$ is equal to $l$. In the diagram above, suppose that the red, dotted arc $ADB$ and the upper part of the circle both have length $l$. Note that the region bounded by $ADBC$ (i.e.: the pink area) has the same perimeter length as the circle.
While I'm not sure that this will perfectly solve the matter for arbitrary $l$, it's easy to get this in special cases, in a way that may point the way to a general solution (in particular, by continuity) : build an isosceles triangle on $AB$ with the other two sides both equal to the radius of the appropriate circle, and consider the 'ice-cream-cone' with your curve and that the two radii, and piece together however many of these it takes to form a circle (note that this only works for arcs that are integral divisions of the appropriate circle!). The isoperimetric inequality then says that the total area-maximizing curve is a circle, which implies that the section of curve between $AB$ that maximizes the wedge area is a circular arc, and since the area of the triangle is constant you can just subtract it out...
I had thought about this, but I wasn't really convinced because it wasn't obvious to me why the solution to the unconstrained problem should be related to the solution for the constrained problem. In effect, the assertion seems to be that, look, we can always draw $AB$ as a chord in the circle, so that must be the answer... but I'm not convinced. – Zhen Lin Feb 18 '11 at 20:42