Theorem: Nine Point Circle Theorem

Let $\triangle ABC$ be a triangle.
These $9$ points:
:the feet of the altitudes of $\triangle ABC$
:the midpoints of the sides of $\triangle ABC$
:the midpoints of the lines from the vertices of $\triangle ABC$ to the orthocenter $H$ of $\triangle ABC$
all lie on the circumference of a circle.
The center $M$ lies on the Euler line of $\triangle ABC$, at the midpoint between the orthocenter $H$ and the circumcenter $O$.
:500px

Proof:
Let the altitudes of $\triangle ABC$ be $AD$, $BE$ and $CF$.
Let $H$ be the orthocenter of $\triangle ABC$.
Let $X$, $Y$ and $Z$ bisect $AH$, $BH$ and $CH$, respectively.
Let $A_m$, $B_m$, and $C_m$ bisect $BC$, $AC$, and $AB$, respectively.
:450px
{{begin-eqn}}
{{eqn | l = \triangle AHC
      | o = \sim
      | r = \triangle B_mZC
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{eqn | ll = \leadsto
      | l = AXHD
      | o = \parallel
      | r = B_mZ
      | c =
}}
{{eqn | l = \triangle AHB
      | o = \sim
      | r = \triangle C_mYB
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{eqn | ll = \leadsto
      | l = AXHD
      | o = \parallel
      | r = C_mY
      | c =
}}
{{eqn | l = B_mZ
      | o = \parallel
      | r = C_mY
      | c = Parallelism is Transitive Relation
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \triangle YZH
      | o = \sim
      | r = \triangle BCH
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{eqn | ll = \leadsto
      | l = YZ
      | o = \parallel
      | r = BDC
      | c =
}}
{{eqn | l = \triangle AC_mB_m
      | o = \sim
      | r = \triangle ABC
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{eqn | ll = \leadsto
      | l = C_mB_m
      | o = \parallel
      | r = BDC
      | c =
}}
{{eqn | l = C_mB_m
      | o = \parallel
      | r = YZ
      | c = Parallelism is Transitive Relation
}}
{{end-eqn}}
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$\Box B_mZYC_m$ is a parallelogram.
{{begin-eqn}}
{{eqn | l = AXHD
      | o = \perp
      | r = BC
      | c = {{Defof|Altitude of Triangle}}
}}
{{eqn | l = ZY
      | o = \parallel
      | r = BC
      | c = above
}}
{{eqn | ll = \leadsto
      | l = ZY
      | o = \perp
      | r = AXHD
      | c =
}}
{{eqn | l = B_mZ
      | o = \parallel
      | r = AXHD
      | c = above
}}
{{eqn | ll = \leadsto
      | l = B_mZ
      | o = \perp
      | r = ZY
      | c =
}}
{{eqn | l = \angle B_mZY
      | r = 90 \degrees
      | c = {{Defof|Perpendicular}}
}}
{{end-eqn}}
By Parallelogram with One Right Angle is Rectangle:
:$\Box B_mZYC_m$ is a rectangle.
{{begin-eqn}}
{{eqn | l = B_mY
      | r = C_mZ
      | c = Diagonals of Rectangle are Equal
}}
{{end-eqn}}
Draw the circle with $B_mY$ as diameter.
By Thales' Theorem/Converse:
:$Z$ and $C_m$ are also on the circle.
Since $ZC_m = Y B_m$:
:$ZC_m$ is a diameter of the circle.
And {{mutatis}}:
:$XA_m$ is a diameter of the circle.
Using diameter $B_mY$, by Thales' Theorem/Converse:
:$E$ is on the circle.
Using diameter $C_mZ$, by Thales' Theorem/Converse:
:$F$ is on the circle.
Using diameter $A_mX$, by Thales' Theorem/Converse:
:$D$ is on the circle.
The result follows.
{{qed}}