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 · AB = 0).

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I am struggling to show how CE and AB are perpendicular.

  • $\begingroup$ There is a theorem saying that the three "orthogonal-to-opposite-edge" lines intersect at a mutual point. Given this theorem, what could possibly be the line from C orthogonal to AB? $\endgroup$ – uniquesolution May 23 '16 at 8:23
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    $\begingroup$ @uniquesolution I think the asker is asking for a proof of that well known theorem in basic Euclidean geometry but using vectors $\endgroup$ – DonAntonio May 23 '16 at 8:25
  • $\begingroup$ For example: $(CD,AB) = (CA,AC)+(CA,CB)+(AD,AC) = $ $(CA,AC)+(CA,CA) + (CA,AD) + (AD,AC)$, where $(u,v)$ is a scalar product of $u,v$. $\endgroup$ – Abstraction May 23 '16 at 8:29


$$\vec{AB}=u\;,\;\;\vec{AC}=v\implies \vec{BC}=u-v$$

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


$$\vec{CD}\cdot\vec{AB}=(tu+(s-1)v)\cdot u=t\left\|u\right\|^2+(s-1)u\cdot v=$$

$$=t\left\|u\right\|^2+\overbrace{\left(tu\cdot v-t\left\|u\right\|^2+s\left\|v\right\|^2\right)}^{=su\cdot v\;(1)}-u\cdot v=(t-1)u\cdot v+s\left\|v\right\|^2\;\;\color{red}{(3)}$$

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 $\;\vec{CD}\perp\vec{AB}\;\;\;\square$


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