Solving perpendicularity problem not using scalar product of vector Given quadrilateral $ABCD$ and $O=AC\cap BD$. Let $H$, $K$, respectively, be the orthocenters of $\triangle AOD$, $\triangle BOC$; $M$, $N$, respectively, be the midpoints of $AB$, $CD$. Prove that $MN\perp HK$.
One solution for this question is using "scalar product", which means proving $\vec{MN}.\vec{HK}=0$ and it's not very difficult!
However, I'm looking forward to solving this problem not by using anything related to vector. It means that we prove $MN\perp HK$ just using the properties in triangles (similarity, equality, etc.), quadrilaterals and circles. Many thanks!
 A: Let $E$ be the midpoint of $AC$. Since $EN$ is a middle line of $\triangle CAD$, we have $EN\parallel AD$ and $EN=AD/2$. Similarly, $EM\parallel CB$ and $EM=CB/2$. In particular, $EN:EM=AD:CB$. Since $OH\perp AD$ and $OK\perp CB$, we get $OH\perp EN$ and $OK\perp EM$. Also $OH=AD\cdot|\cot\angle AOD|$ and $OK=CB\cdot |\cot\angle COB|$, so $OH:OK=AD:CB$ as $\angle AOD=\angle COB$.
Since $\angle HOK$ and $\angle NEM$ are angles with perpendicular rays we have two possibilities: $\angle HOK=\angle NEM$ or $\angle HOK+\angle NEM=180^\circ$. I will discuss that the first option holds having in mind the following picture (discussions are similar in other cases):

Note that $\angle NEM=$ $180^\circ-\angle NEC+\angle MEA=$ $180^\circ-\angle DAC+\angle BCA=$ $\angle ADC+\angle DCA+\angle BCA=$ $\angle ADC+\angle BCD$ and $\angle HOK>\angle AOB=\angle DOC$, so $\angle NEM+\angle HOK> \angle ADC+\angle BCD+\angle DOC= 180^\circ+\angle ADB+\angle BCA>180^\circ$, so the second option doesnt hold.
So $EN:EM=OH:OK$ and $\angle NEM=\angle HOK$, hence $\triangle NEM\sim\triangle HOK$. Therefore, $EN\perp OH$ and $EM\perp OK$ imply $NM\perp HK$.
