The figure shows an arbelos ABC (AB, BC, and AC are semicircles
of centers O_{1}, O_{2}, and O). The semicircles
of diameters AO_{2} (center O_{3}) and CO_{1} (center O_{4})meet at E. BD is perpendicular to AC,
AD intersects semicircles O_{1} and O_{3} at A_{1}
and A_{3}, respectively, CD intersects semicircles O_{2} and O_{4} at C_{2} and
C_{4}, respectively. Prove that (1) B, E, D are
collinear points; (2) A_{1}C_{2} is
the common tangent to semicircles O_{1} and O_{2} at A_{1} and C_{2},
respectively; (3) A_{3}C_{4} is the common tangent to
semicircles O_{3}
and O_{4} at A_{3} and C_{4}, respectively;
(4) A_{3} is the midpoint of DA_{1} and C_{4}
is the midpoint of DC_{2}; (5) M is the midpoint of A_{3}C_{4}; (6) A_{3}C_{4}
// A_{1}A_{2}; (7) A_{1}C_{2}
= BD = 2A_{3}C_{4}. See also:
Artwork Problem
1071.
