A geometry problem - proving collinearity Let a triangle $ABC$. $M$ is a point inside triangle. Construct the line through M perpendicular to $MA, MB, MC$ and intersect $BC, CA, AB $ at $A_0,B_0,C_0$ respectively. Prove that $A_0,B_0,C_0$ are collinear.
I really don't know where to start. Menelaus? Inversion?


Let $M$ be the origin and consider the three vertices $A$, $B$, $C$ as vectors $a$, $b$, $c$. You get the vector $a'=(1-t)b+t c\>$ representing $A_0$ by solving
$$\bigl((1-t)b+tc\bigr)\cdot a=0$$
for $t\in{\mathbb R}$ and obtain
$$a'={(a\cdot b)c-(a\cdot c)b\over a\cdot(b-c)}\ .$$
Using analogy for $b'$ and $c'$ it follows that
$$\bigl(a\cdot(b-c)\bigr) a'+\bigl(b\cdot(c-a)\bigr) b'+\bigl(c\cdot(a-b)\bigr) c'=0\ .$$
But $\lambda a'+\mu b'+\nu c'=0$ with $\lambda+\mu+\nu=0$ means that $a'$, $b'$, and $c'$ are collinear.
