Question

$P$ is any point on the circumcircle of $△ABC$ other than the vertices. $H$ is the orthocenter of $△ABC,M$ is the mid-point of $PH$ and $D$ is the mid-point of $BC$. Then

• A. $AP$ is opposite side of $DM$
• B. $DM$ is parallel to $AP$
• C. $DM$ is perpendicular to $AP$
• D. None of these

Answer 1

The correct option is C $DM$ is perpendicular to $AP$

Let $S$ be the circumcentre of $△ABC.$

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ and $\overrightarrow{p}$ be the orthocentre of vectors of $A,B,C$ and $P$ respectively with referentre to origin $S.$

$\overrightarrow{SA}=\overrightarrow{a},\overrightarrow{SB}=\overrightarrow{b},\overrightarrow{SC}=\overrightarrow{c},\overrightarrow{SP}=\overrightarrow{p}$ and $\left|\overrightarrow{p}\right|=\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=R=$ circumradius.

Now, $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}$

$=\overrightarrow{SA}+\overrightarrow{2SD}=\overrightarrow{SA}+\overrightarrow{AH}=\overrightarrow{SH}$

$(\because H.G:GS=2:1$ where $G$ is centroid $)$

Position vector of $H=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$

So, position vector of $M=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{p}}{2}$

$\overrightarrow{DM}=p.v$ of $M-p.v.$ of $D=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{p}}{2}-\frac{\overrightarrow{b}+\overrightarrow{c}}{2}=\frac{\overrightarrow{a}+\overrightarrow{p}}{2}$

Now $\overrightarrow{DM}.\overrightarrow{PA}=\left(\frac{\overrightarrow{a}+\overrightarrow{p}}{2}\right).\left(\frac{\overrightarrow{a}-\overrightarrow{p}}{2}\right)=\frac{{\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{p}\right|}^2}{2}=0\Rightarrow \overrightarrow{DM}\perp \overrightarrow{AP}$