Be an ABC triangle and a point P of your plane. The perpendiculars to PA, PB and PC, traced by P, intersect the BC, CA and AB sides at three points, A ', B' and C '. Prove that these points are in a straight line.
Can someone explain me how to do this question? I just found that $\angle APB'=\angle A'PB\\ \angle B'PC=\angle BPC'$ My first idea was to use Analytic Geometry, but I don't know what I have to do... Thanks for antetion. If someone has a solution by eucliidean geometry, I will love it!
There should be probably a nice Euclidean solution, however, I would like to show you a short vector one.
Let $P$ be the origin and points $A,B,C$ have coordinates $\mathbf{a},\mathbf{b},\mathbf{c}$. Let's say that $BA'/BC=\alpha$, then the coordinate of $A'$: $$ \mathbf{a'}=\mathbf{b}+\alpha(\mathbf c-\mathbf b). $$ By construction, we want $\mathbf{a}\cdot\mathbf{a'}=0$, so $$ (\mathbf{b}+\alpha(\mathbf c-\mathbf b))\mathbf a=0,\\ \alpha=-\frac{\mathbf a\mathbf b}{\mathbf a\mathbf c-\mathbf a\mathbf b}. $$ Then, we can find that $$ \frac{BA'}{A'C} = \frac{\alpha}{1-\alpha} = -\frac{\mathbf a\mathbf b}{\mathbf a\mathbf c}. $$ By analogy, $CB'/B'A=-\mathbf b\mathbf c/\mathbf a\mathbf b$ and $AC'/C'B=-\mathbf a\mathbf c/\mathbf b\mathbf c$.
Finally, since $$ \frac{BA'}{A'C}\frac{CB'}{B'A}\frac{AC'}{C'B}=-1, $$ by Menelaus theorem we can conclude that $A',B',C'$ are colinear.