Draw lines, parallel to the sides of the given rectangle, through the vertices of the triangle and consider the least rectangle $ABCD$ thus formed which still contains the triangle. In the general case, one vertex of the triangle is situated at $A$, while the other two are, say $X$ on the side $BC$ and $Y$ on the side $DC$ (with the possibility that they are even situated at $B, C$ or $D$). The fourth point $Z$ is inside $\triangle AXY$ . Denote by $X'$ the point on side $AD$ such that $XX'\parallel AB$ and by $Y'$ the point on side $AB$ such that $YY'\parallel AD$. Also denote by $O$ the meeting point of $XX'$ and $YY'$. The(some of them possibly degenerated)triangles $AXX',AYY',XOY$ together cover the triangle $AXY$ . Say $Z$ lies in $\triangle XOY$ ; then its symmetric $Z'$ with respect to the midpoint of $XY$ is contained in the rectangle $ABCD$. But the quadrilateral $AXZ'Y$ is convex, and also contained in the rectangle $ABCD$, thus by a well-known property of plane convex bodies, its perimeter is less than that of $ABCD$, in turn less thanthat of the original rectangle. Since $ZX = Z'Y$ and $ZY = Z'X$, the perimeter of $AXZY$ is equal to that of $AXZ'Y$ , and we are done. The other cases, when $Z$ lies in $\triangle AXX'$ or $\triangle AYY'$ (or both), are treated in a completely similar way.