Respuesta :
The theorem:
-Given triangle ABC. Let point D be the midpoint of side AB, point E be the midpoint of side AC, and point F be the midpoint of side BC (Note triangle DEF is the medial triangle).
-Let point J be the foot of the altitude from point C to side AB, point K be the foot of the altitude from point B to side AC, and point L be the foot of the altitude from point A to side BC (Note triangle JKL is the orthic triangle).
-And if point H is the orthocenter of triangle ABC, let point M be the midpoint of HB, point N be the midpoint of HA, and point P be the midpoint of HC (Note triangle MNP is the orthocenter mid-segment triangle).
-Prove points D, E, F, J, K, L, M, N, and P lie on a common circle (called the nine-point circle).
The proof:
-In triangle ABC, since D and E are midpoints of sides AB and AC (by construction), then segment DE is parallel to side BC.
-In triangle BCH, since M and P are midpoints of sides BH and HC (by construction), then segment MP is parallel to side BC.
-Since segment DE is parallel to side BC, and segment MP is parallel to side BC, then segments DE and MP are parallel to each other.
-In triangle BAH, since M and D are midpoints of sides BH and BA (by construction), then segment MD is parallel to side HA.
-In triangle CAH, since E and P are midpoints of sides AC and HC (by construction), then segment EP is parallel to side HA.
-Since segment MD is parallel to side HA, and segment EP is parallel to side HA, then segments MD and EP are parallel to each other.
-Since segment MD is parallel to segment HA, and segment HA lies on segment AL, then segment MD is parallel to segment AL.
-Since segment AL is perpendicular to side BC by construction (AL is the altitude from point A to side BC), and side BC is parallel to segment DE, then segment AL is perpendicular to segment DE.
-Since segment MD is parallel to segment AL, and segment AL is perpendicular to segment DE, then segment MD is perpendicular to segment DE.
-Thus, we have that quadrilateral DEPM is a rectangle. Since the opposite angles of this quadrilateral are supplementary, it follows that the quadrilateral can be inscribed in a circle.
-Similarly, quadrilateral DNPF is a rectangle, and it can be inscribed in a circle.
-Therefore, points D, N, E, P, F, and M are on a common circle, with one diameter of the circle being segment DP, since this segment is a diagonal of both rectangles. The center of this circle, then, is the midpoint of segment DP. Let this center be point O.
-Now, since segment AL is an altitude, angle NLF is a right angle. But segment NF is also a diameter of our circle (it is a diagonal of rectangle DNPF), so it follows that point L must lie on this circle.
-Similarly, points J and K are on the circle.
-Therefore, all nine points D, E, F, J, K, L, M, N, and P lie on the same circle, called the nine-point circle
