How to prove that:$ BC^2= 3CM^2 + AC^2 $from this triangle problem?

Question

In the triangle $\triangle ABC$, angle $\angle A$ is larger than angle $\angle B$.

We choose points $M$ and $N$ at $AB$ such that $AM=MN=NB$.

How to prove that: $BC^2= 3CM^2 + AC^2$?

Which theorem(s) should I use to prove this problem?

Thanks

Answer 1

Points $A,M,N$ and $B$ are equally spaced on $(AB)$. Set a coordinate system with origin at $A$ and $(AB)$ as first axis.

Then the coordinates of $A,M,N$ and $B$ will be :

$$ A(0,0) \ M(t,0) \ N(2t,0) \ B(3t,0) $$

for some constant $t$. Now, if we denote by $(x,y)$ the coordinates of $C$, one has

$$ \begin{array}{lcl} BC^2 &=& (x-3t)^2+y^2=x^2+y^2-6xt+9t^2 \\ 3CM^2 &=& 3((x-t)^2+y^2)=3x^2+3y^2-6xt+3t^2 \\ AC^2 &=& (x-0)^2+y^2=x^2+y^2 \\ \end{array} $$

and hence

$$ BC^2-(3CM^2+AC^2)=-3x^2-3y^2+6t^2 $$

so your equality holds iff $x^2+y^2=2t^2$. You probably forgot a condition on $C$