[br][br]Let [i]ABC[/i] be any triangle, and [i]A´[/i] , [i]B´[/i] and [i]C´[/i] be arbitrary points on lines [i]BC[/i], [i]AC[/i], and [i]AB[/i][br]respectively. Then the 3 [url=https://en.wikipedia.org/wiki/Circumcircle]circumcircles[/url] to triangles [i]AB´C´[/i], [i]A´BC´[/i], and [i]A´B´C [/i]intersect in a unique point[i] M, [/i]called Miquel's point. [br][br]Moreover, if the points [i]A´[/i], [i]B´ [/i]and [i]C´[/i] are on the segments [i]BC[/i], [i]AC[/i], and [i]AB[/i] , then the three angles [i]MA´B[/i], [i]MB´C[/i] and [i]MC´A[/i] are equal, and (of course also) the three supplementary angles [i]MA´C[/i], [i]MB´A[/i] and [i]MC´B[/i].[br][br]Check what happens if the points [i]A´[/i], [i]B´[/i] and [i]C´[/i]  are not on the segments [i]BC[/i], [i]AC[/i], and [i]AB.[/i][br][br]If the points [i]A´[/i], [i]B´[/i] and [i]C´[/i] are colinear, then the point [i]M[/i] belongs to the circumcircle of the triangle[br]ABC. And vice-versa. To check this, the line B’C’ has been plotted (in red). Move the point A’ until it is on this line and check what happens. [br][br][br]