Let the unit area rectangle $ABCD$ have the incenter of the triangle determined by vertices $A, B, C$ as $O$. Consider the rectangle with $O$ and $D$ as opposite vertices, and the other two vertices on the perimeter of the original rectangle. What is the area of this rectangle?
We will show that the rectangle with vertex $O$ is cut by the diagonal $AC$ into a triangle - which is shaded in the diagram - whose area is equal to the part of the $\triangle ACD$ that protrudes from the rectangle. (In the diagram, the two dotted right-angled triangles are shown.) This implies that the areas of the two figures, the rectangle with vertex $O$ and the $\triangle ACD$, are equal, so the area of the rectangle in question is $1/2$.

# $1986-12-447-1$. eps 

Indeed, the radius of the inscribed circle of $\triangle ABC$ perpendicular to the diagonal $AC$ cuts the shaded triangle into two right-angled triangles, each with a leg of length $r$. Due to the equal vertex angles, one of these is congruent to the triangle at vertex $C$, and the other is congruent to the triangle at vertex $A$, both with a leg of length $r$.