Prove that the perpendiculars from P,Q,R to BC,CA,AB respectively are also concurrent.

Let $ABC$ and $PQR$ be any two triangles in the same plane. Assume that the perpendiculars from the points $A,B,C$ to the sides $QR,RP,PQ$, respectively, are concurrent. Prove that the perpendiculars from $P,Q,R$ to $BC,CA,AB$ respectively are also concurrent.

I was suggested to prove this using vectors. But I have no idea how to use them...

Solution 1:

Such two triangles are called orthologic triangles. We claim that the perpendiculars from $ABC$ to $PQR$ concur iff $$MA\cdot QR+MB\cdot RP+MC\cdot PQ=0$$ for every point $M$ in the plane (where each $XY$ denotes the vector $\vec{XY}$).

First, notice that $f(M)=MA\cdot QR+MB\cdot RP+MC\cdot PQ=0$ is constant with respect to $M$, as $$f(M)-f(N)=MN\cdot (QR+RP+PQ)=MN\cdot 0=0$$ Now, if we take $M$ to be the intersection of the perpendiculars from $A$ to $QR$ and $B$ to $RP$, we get that $MA\cdot QR+MB\cdot RP=0$. Then $$f(M)=MC\cdot PQ=0$$ iff $MC$ is perpendicular to $PQ$, or equivalently if the three perpendiculars concur at $M$.

Now, it is not hard to see that the vector condition is symmetric with respect to switching $ABC$ and $PQR$. Indeed, if we take $M$ to be the origin, $$ MA\cdot QR+MB\cdot RP+MC\cdot PQ=0 $$ $$ \Leftrightarrow A\cdot Q-A\cdot R+B\cdot R-B\cdot P+C\cdot P-C\cdot Q=0 $$ $$ \Leftrightarrow MP\cdot BC+MQ\cdot CA+MR\cdot AB=0 $$ So the two concurrences are equivalent.