P is ant point on the circum-circle of ΔABC other than the vertices. H is the orthocentre of ΔABC, M is the mid-point of PH and D is the mid-point of BC. Then,

DM is perpendicular to AP

Let the origin O be the circumcentre of ΔABC and let the position vectors of vertices A, B and C be $\vec a,\vec b$ and $\vec c$ respectively. Then, the position vector of its centroid G is $\frac{\vec a+\vec b+\vec c}{3}$. Since the centriod G divides the segment joining circumcentre O and orthocentre H in the ratio 1 : 2. So, the position vector of H is $\vec a+\vec b+\vec c$. Let $\vec r$ be the position vector of point P on the circumcircle. Then, the position vector of M is $\frac{\vec a+\vec b+\vec c+\vec r}{2}$. The position vector of D is $\frac{\vec b+\vec c}{2}$.

$∴\vec{DM}=\frac{\vec a+\vec r}{2}$

Now, $\vec{DM}.\vec{AP}=(\frac{\vec a+\vec r}{2}).(\vec r-\vec a)=\frac{1}{2}(|\vec r|^2-|\vec a|^2)=0$

Hence, $DM ⊥AP$.