Problem 1049

Solution. Triangle $CDE$ is isosceles, and $AD$ is the height to its base. Therefore, $AD$ is the bisector of triangle $CDE$, and angles $ADE$ and $ADC$ are equal. Angles $ADE$ and $BAD$ are equal as alternate interior angles when parallel lines $AB$ and $DE$ are intersected by the transversal $AD$. Thus, angles $ADC$ and $BAD$ are equal. Since lines $BC$ and $AD$ are perpendicular to the same line, they are parallel, and $ABCD$ is an isosceles trapezoid, from which $AB = CD = DE$. Therefore, $ABDE$ is a parallelogram. Let $O$ be the point of intersection of its diagonals $AD$ and $BE$, then $AO = OD, BO = OE$.

Let $X$ be the point of intersection of the line passing through $B$ parallel to $CE$ and the line passing through $E$ parallel to $BC$. Then $BCEX$ is a parallelogram. Point $O$ is the midpoint of its diagonal $BE$, so it is also the midpoint of diagonal $CX$. Therefore, the diagonals $AD$ and $CX$ of quadrilateral $ACDX$ are bisected by their point of intersection. Thus, $ACDX$ is a parallelogram, meaning $AX$ is parallel to $CD$, and all three lines specified in the problem

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intersect at one point.