<s>[INST] The existence of a point in a geometric configuration Consider $ABC$ a right triangle in $A$ and let $M$ be the middle of the height from the vertex $A$. Construct $B'=f(B)$ and $C'=g(C)$, where $f$ is the reflection about line $CM$ and $g$ is the reflection about line $BM$. Decide when do lines $BC'$ and $CB'$ intersect? When $BC'\cap CB'={E}$, prove that $BE+CE=\frac{5}{3}BC$.
I tried to find a length or an angle that takes some values when the lines intersect each other and other values when the lines do not intersect. I do not get anything concrete.
For the second part I tried to apply bisector theorem but I didn't obtain anything.
 [/INST] Let $D$ be the feet of altitude from $A$ to $BC$.  Then $AD=\frac{bc}a$, $DB=\frac{c^2}a$ so $\tan\angle MBC=\frac{b}{2c}$ and hence
$$
\tan\angle C'BC=\frac{4bc}{4c^2-b^2},
$$
and similarly
$$
\tan\angle B'CB=\frac{4bc}{4b^2-c^2}.
$$
The two lines $C'B$ and $B'C$ does not intersects iff $\angle C'BC+\angle B'CB=180^\circ$, i.e., $4b^2-c^2=-(4c^2-b^2)$, which is impossible.  So the two lines always intersect.
Now use $\sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}$ and similar to calculate
$$
\sin\angle ECB=\frac{4bc}{4b^2+c^2},\quad
\sin\angle EBC=\frac{4bc}{4c^2+b^2},\quad
\sin\angle BEC=\frac{12a^2bc}{(4b^2+c^2)(4c^2+b^2)}.
$$
So sine rule gives
\begin{align*}
BE+CE&=\frac{a}{\sin\angle BEC}\cdot(\sin\angle ECB+\sin\angle EBC)\\
&=a\cdot\frac{(4b^2+c^2)(4c^2+b^2)}{12a^2bc}\left(\frac{4bc}{4b^2+c^2}+\frac{4bc}{4c^2+b^2}\right)\\
&=\frac{4abc}{12a^2bc}((4b^2+c^2)+(4c^2+b^2))\\
&=\frac{5(b^2+c^2)}{3a}=\frac53a
\end{align*}
as desired.
</s>