Let A, B and C be non-colinear points in a plane. Form the triangle ABC, and from each of A and B, draw the line that is orthogonal to the opposite side. These two lines meet at a point D. Now let E be the intersection of AB and the line through C and D. Prove that CE is orthogonal to AB (Hint: we want to show that CD ยท AB = 0).
I am struggling to show how CE and AB are perpendicular.
Best Answer
Denote
$$\vec{AB}=u\;,\;\;\vec{AC}=v\implies \vec{BC}=u-v$$
We then can write $\;\vec{AD}=tu+sv\;,\;\;t,s\in\Bbb R\;,\;\;\vec{BD}=mu+nv\,,\,\,m,n\in\Bbb R$ , and we're given
$$\vec{AD}\perp \vec{BC}\implies(tu+sv)\cdot(u-v)=t\left\|u\right\|^2-s\left\|v\right\|^2+(s-t)u\cdot v=0\;\;\color{red}{(1)}$$
and also
$$\vec{BD}\perp\vec{AC}\implies(mu+nv)\cdot v=mu\cdot v+n\left\|v\right\|^2=0\;\;\color{red}{(2)}$$
so that we get
$$\vec{CD}=-v+tu+sv\implies$$
$$\vec{CD}\cdot\vec{AB}=(tu+(s-1)v)\cdot u=t\left\|u\right\|^2+(s-1)u\cdot v=$$
$$=t\left\|u\right\|^2+\overbrace{\left(tu\cdot v-t\left\|u\right\|^2+s\left\|v\right\|^2\right)}^{=su\cdot v\;(1)}-u\cdot v=(t-1)u\cdot v+s\left\|v\right\|^2\;\;\color{red}{(3)}$$
But we also have that
$$\vec{BD}=\vec{BA}+\vec{AD}=-u+tu+sv=(t-1)u+sv\implies\text{we can write as follows}\;(2):$$
$$0=\vec{BD}\cdot\vec{AC}=((t-1)u+sv)\cdot v=s\left\|v\right\|^2+(t-1)u\cdot v=\;\color{red}{(3)}\;\;!!$$
and this means $\;\vec{CD}\perp\vec{AB}\;\;\;\square$