Two altitudes of a triangle $ABC$ meet at $H$, as shown. Relative to some origin $O$ the position vectors of $A$, $B$, $C$ and $H$ are $a$,$b$,$c$ and $h$. Show that $$ (h-a)\cdot(b-c)=(h-b)\cdot(c-a)=0 $$ Deduce that $(h-c)\cdot(b-a)=0$ and hence that the three altitudes of a triangle are concurrent.
I tried expanding the brackets but it confused me and I couldn't get anything meaningful from it, I also tried substituting different vectors to get to the same vector position but I couldn't derive anything meaningful. I also tried putting them as a dot product as shown below:
$$ (h-a).(b-c)=\sqrt{(h-a)^2(b-c)^2}\cos(x) $$ $$ \cos(x)=1 $$ $$ x=0° $$
However, that means they are parallel which is not the case or they wouldn't equal $0$.
Anymore advice you could give me?
Using the distributive property render
$(h-a)\cdot(b-c)=h\cdot b-a\cdot b -h\cdot c+a\cdot c$.
Do the same for
$(h-b)\cdot(c-a)=...$
$(h-c)\cdot(a-b)=...$
And add them up.
$(h-a)\cdot(b-c)+(h-b)\cdot(c-a)+(h-c)\cdot(a-b)=...$ (hint: you might find yourself canceling terms.)
Thereby if two of the terms in the sum are zero, then the third term is ... .