A quadrilateral inscribed in a rectangle Given a rectangle $ABCD$ in which there is an inscribed quadrilateral $XYZT$, with exactly one vertex on each side of the rectangle, how could I prove that the perimeter of the inscribed quadrilateral is larger then $2|AC|$ (two diagonals)?
I tried to use the triangle inequality, but I can't find the right way to do it.

Let’s suppose that $X$, $Y$, $Z$, $T$ are in $AB$, $BC$, $CD$, $DA$, respectively. Construct the reflection of $X$ through $AD$ $X_1$, the reflection of $X$ through $BC$ $X_2$, and the reflection of $X_1$ through $CD$ $X_3$. Your diagram should look like this.

Now, we have $$XY+YZ+ZT+TX$$ $$=X_2Y+YZ+ZT+TX_1$$ $$\ge X_2Z+ZX_1$$ $$=X_2Z+ZX_3$$ $$\ge X_3X_2.$$ However, since $X_1X_2=2AB$, $X_1X_3=2AD$, $\angle X_3X_1X_2=\angle DAB=90^\circ$, $$\bigtriangleup X_1X_2X_3\sim\bigtriangleup ABD,$$ so that $$X_3X_2=2AC,$$ and $XY+YZ+ZT+TX\ge 2AC$, as we wanted. $\blacksquare$
