<s>[INST] How to prove that:$ BC^2= 3CM^2 + AC^2 $from this triangle problem? 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
 [/INST] 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 $C$
</s>