Miquel Point
Idea · Intermediate
Geometry
Given triangle and points , , and on , , and , respectively, the circumcircles of triangles , , and have a common point , called the Miquel point of the configuration.


Let be the second intersection of the circumcircles of and .
Since is cyclic,
Since is cyclic,
Therefore,
Because are collinear, these equal angles imply that are cyclic. Thus also lies on the circumcircle of .
Unlike the circumcenter or incenter, the Miquel point generally depends on the chosen points , , and , not just on triangle .
Consider four lines in general position, with no two parallel and no three concurrent.
Each choice of three lines determines a triangle, giving four triangles in total.
The circumcircles of these four triangles pass through a common point.
This is known as the Four-Line Miquel Theorem.

Let be the second intersection of and .

Therefore, is cyclic.

Therefore, is cyclic. Hence lies on the remaining two circumcircles, and .
Let be an acute triangle. Let , , and be the feet of the altitudes from , , and , respectively, and let be the orthocenter. Prove that the Miquel point of , , and is .

Thus is cyclic.
Similarly, and are cyclic.
Therefore, is the Miquel point.
Let the incircle of triangle touch , , and at , , and , respectively. Let be the incenter. Prove that the Miquel point of , , and is .

Thus is cyclic. Similarly, and are cyclic.
Therefore, is the Miquel point.