In a given trapezium
with
parallel to
and
, the line
bisects the angle
. The line through
parallel to
meets the segments
and
in
and
, respectively. Let
be the circumcenter of the triangle
. Suppose that
. Prove the equality
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In a given trapezium $ABCD$ with $AB$ parallel to $CD$ and $AB > CD$, the line $BD$ bisects the angle $\angle ADC$. The line through $C$ parallel to $AD$ meets the segments $BD$ and $AB$ in $E$ and $F$, respectively. Let $O$ be the circumcenter of the triangle $BEF$. Suppose that $\angle ACO = 60^{\circ}$. Prove the equality
$$CF = AF + FO .$$