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? enter image description here

  • $\begingroup$ In what kind of school is this a "homework"? It looks more difficult than what can be assigned in an ordinary class. $\endgroup$
    – zyx
    Jun 11, 2013 at 2:33

2 Answers 2


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.


Let $A(0,a),B(b,0),C(c,0),M(x_1,y_1)$

$k_{AM}=\dfrac{y_1-a}{x_1},A_0M: y-y_1=\dfrac{x_1}{a-y_1}(x-x_1) \to A_0 \left(\dfrac{x_1^2+y_1^2-ay_1}{x_1},0\right)$

$k_{BM}=\dfrac{y_1}{x_1-b},B_0M:y-y_1=-\dfrac{x_1-b}{y_1}(x-x_1),AC:y=-\dfrac{a}{c}(x-c) \to B_0 \left(\dfrac{c(y_1^2+x_1^2-ay_1-bx_1)}{cx_1-ay_1-bc},-\dfrac{a(y_1^2+x_1^2-cx_1-bx_1-bc)}{cx_1-ay_1-bc} \right)$

$k_{cM}=\dfrac{y_1}{x_1-c},C_0M:y-y_1=-\dfrac{x_1-c}{y_1}(x-x_1),AB:y=-\dfrac{a}{b}(x-b) \to C_0 \left(\dfrac{-b(y_1^2+x_1^2-ay_1-cx_1)}{bx_1-ay_1-bc},-\dfrac{a(y_1^2+x_1^2-bx_1-cx_1-bc)}{bx_1-ay_1-bc} \right)$



$k_{A_0B_0}=k_{A_0C_0} \implies A_0,B_0 ,C_0 $ co-line .

BTW, the mid points of $AA_0,BB_0,CC_0$ are also co-line.


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