# 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!

• You say: One solution for this question is using "scalar product", which means proving $\vec{MN}.\vec{HK}=0$ and it's not very difficult! Can you show it please?! – Aqua Apr 4 at 15:53
• My solution using "scalar product": First, prove $\overrightarrow{MN}=\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{BD} \right)$. Then, let $A',C'$ be the projections of $A,C$ on $BD$; $B',D'$ be the projections of $B,D$ on $AC$. Then, we have $\overrightarrow{HK}.\overrightarrow{AC}=\overrightarrow{B'D'}.\overrightarrow{AC}=\overrightarrow{DB}.\overrightarrow{AC}$ and $\overrightarrow{HK}.\overrightarrow{BD}=\overrightarrow{A'C'}.\overrightarrow{BD}=\overrightarrow{AC}.\overrightarrow{BD}$. Therefore, we can prove $\overrightarrow{MN}.\overrightarrow{HK}=0$. – Martin Tr Apr 10 at 1:16

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 $$NM\perp HK$$.