parmenides51 wrote: Let $AE$ be a diameter of the circumcircle of triangle $ABC$. Join $E$ to the orthocentre, $H$, of $\triangle ABC$ and extend $EH$ to meet the circle again at $D$. Prove that the nine point circle of $\triangle ABC$ passes through the midpoint of $HD$. Note. The nine point circle of a triangle is a circle that passes through the midpoints of the sides, the feet of the altitudes and the midpoints of the line segments that join the orthocentre to the vertices.

Hint

Generalisation: The Nine-Point Circle passes through the mid-point of the line segment joining the orthocenter to any point on the circumcircle

proof

We see that: NPC is image of ($ABC$) through the homothety with center $H$, ratio $k$ = $\dfrac{1}{2}$ But: $D$ $\in$ ($ABC$) then midpoint of $HD$ lies on NPC of $\triangle$ $ABC$