Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
The circumcentres of four triangles are joined to form a quadrilateral. What do you notice about this quadrilateral as the dynamic image changes? Can you prove your conjecture?
The following solution was sent by `kevin295'.
The segments $CD$ and $EF$ are parallel.
$ABDC$ is a cyclic quadrilateral, therefore angle $CDB$ + angle $CAB=180^{\circ}$ (1)
Angle $CAB$ and angle $BAE$ are adjacent angles on a straight line so angle $BAE$ + angle $CAB = 180^{\circ}$ (2)
From (1) and (2) we get angle $CDB$ = angle $BAE$ (3)
Because $ABFE$ is a cyclic quadrilateral, angle $BAE$ + angle $BFE = 180^{\circ}$ (4)
From (3) and (4) we get angle $CDB$ + angle $BFE = 180^{\circ}$
so $CD$ is parallel to $EF$.