Let
be a convex quadrilateral whose sides
and
are not parallel. Suppose that the circles with diameters
and
meet at points
and
inside the quadrilateral. Let
be the circle through the feet of the perpendiculars from
to the lines
and
. Let
be the circle through the feet of the perpendiculars from
to the lines
and
. Prove that the midpoint of the segment
lies on the line through the two intersections of
and
.
Proposed by Carlos Yuzo Shine, Brazil
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Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$.
Proposed by Carlos Yuzo Shine, Brazil