In a $\triangle ABC$, the $m \angle {A} > m \angle {B}$. Points $M$ and $N$ lie on $\overline{AB}$ such that $AM=MN=NB$.

Question

In a $\triangle ABC$, the $m \angle {A} > m \angle {B}$. Points $M$ and $N$ lie on $\overline{AB}$ such that $AM=MN=NB$. Prove $CN^2-CM^2 = \frac{1}{3} \cdot (BC^2-AC^2)$.

It seems I should use Stewart theorem, however, there is the issue that $m \angle {A} > m \angle {B}$ which perhaps has something of concern.

Answer 1

From the given, $CM$ and $CN$ are the medians of the triangles $ACN$ and $BCM$, respectively. Apply the Apollonius' (median) theorem to the two triangles,

$$AC^2+CN^2 = \frac12 AN^2+2CM^2,\>\>\>\>\>BC^2+CM^2 = \frac12 BM^2+2CN^2$$

The given $AM=MN=NB$ leads to $AN=MB$. Then, take the difference of the two equations to obtain

$$CN^2 - CM^2= \frac{1}{3}(BC^2-AC^2)$$

Answer 2

Stewart also works, for this let $a=CB$, $b=CA$, and let $c=AB$ be written in the form $c=3x$, so $AM=MN=NB=x$. Then: $$ \begin{aligned} 3x(CN^2+x\cdot 2x) &= b^2\cdot x+a^2\cdot 2x\ ,\\ 3x(CM^2+2x\cdot x) &= b^2\cdot 2x+a^2\cdot x\ , \end{aligned} $$ and we divide by $x$ and subtract.

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The same argument provides a slight generalization in the situation $$ AM=NC=sc\ ,\ MN=tc\ ,\ s+t+s=1\ . $$ Indeed, we have the same cancellations after $$ \begin{aligned} c(CN^2+s(s+t)\;c^2) &= b^2\cdot sc+a^2\cdot (s+t)c\ ,\\ c(CM^2+(s+t)s\;c^2) &= b^2\cdot (s+t)c+a^2\cdot sc\ , \end{aligned} $$ we divide by $c$, subtract and get $$CN^2-CM^2=t(a^2-b^2)\ .$$ (In the OP we have $s=t=\frac 13$.)