A trapezoid $ABCD$ with bases $AB$ and $CD$ is such that the circumcircle of the triangle $BCD$ intersects the line $AD$ in a point $E$, distinct from $A$ and $D$. Prove that the circumcircle oF the triangle $ABE$ is tangent to the line $BC$.
Supposing (w.l.o.g.) $AB>CD$:
$BCDE$ cyclic means $\measuredangle BED+ \measuredangle BCD=180^\circ$, but $\measuredangle BCD+ \measuredangle ABC=180^\circ$ as well, so $\measuredangle BED=\measuredangle ABC$, done.
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