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 is entirely the
work of Sarah from The Mount School, York (Yr 11). Nice work
I've drawn the triangle and two of the circumcircles and I'll
refer to my drawing :
Now the two circles meet in the two points, D and K.
Now angles DAE and EKD add up to 180 degrees, because DAEK is a
Similarly, FCD and FKD add up to 180 degrees.
Calling the original angles of the triangle A, B and C, angle
(180 - A) + (180 - C) = 360 - (A + C)
So, angle EKF + FBE = 360 - (A + C) + B but A + C = 180 - B,
EKF + FBE = 360 - (180 - B) + B = 180