Given triangle $ABC$ with circumcircle $\Gamma$, let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively, and let the lines $AD$, $BE$, and $CF$ intersect $\Gamma$ again at points $J$, $K$, and $L$, respectively. Show that the area of triangle $JKL$ is at least that of triangle $ABC$.