interior point of a triangle, equal perimeters Let $P$ be an interior point of $\triangle ABC$. Line $AP$ meets $BC$ in $A'$, line $BP$ meets $AC$ in $B'$. Assume that $p(AA'B')=p(BB'A')$. Prove that $p(PA'C)=p(PB'C)$, where $p(XYZ)$ is the perimeter of $\triangle XYZ$.
I'm interested in a geometrical solution, I found this problem on Russian forum, with an ugly computational solution.
 A: HINT AND COMMENT.-You have $$BA'+A'B'+BB'=AA'+A'B'+AB'\iff BA'+BB'=AA'+AB'\cdots (*)$$ Note that this means that the points $A$ and $B$ are in an ellipse whose focal points are the points $A'$ and $B'$ (the sum of the corresponding distances is constant by definition).
You need to prove that (similarly for  $A$ and $B$)
 $$PA'+A'C+PC=PB'+B'C+PC\iff PA'+A'C=PB'+B'C \cdots (**)$$
Let $\left(\dfrac xa \right)^2+\left(\dfrac yb \right)^2=1$ be the ellipse with focal points $B'=(-c,0)$ and $A'=(c,0)$.
Every pair of points  $A$ and $B$ in the ellipse  satisfies $(*)$ but the intersection Q of the lines $\overline{AA'}$ and $\overline{BB'}$ in general does not satisfy $(**)$. You have to calculate $P$ in order $(**)$ becomes true.

So, consider points $A=(x_1,y_1)$ and $B=(x_2,y_2)$ in the ellipse which determines the lines $\overline{AA'}$ and $\overline{BB'}$ whose intersection is the point $P=(x,y)$ with $x=\dfrac{(y_2x_1-y_2c+y_1x_2+y_1c)c}{(y_1+y_2)c+y_1x_2-y_2x_x}$ and $y=\dfrac{y_1(x-c)}{x_1-c}$ or, if you want,
$y=\dfrac{y_2(x+c)}{x_2+c}$.
Besides the lines $\overline{AB'}$ and $\overline{BA'}$ determine the vertix $C$ by intersection. You have now an example of triangle $\triangle{ABC}$ in which you can do the corresponding calculations for suitable $A,B$ and $P$.
I stop here my comment that only aspires to justify in something what you qualify as "ugly computational solution" (I doubt there is a beautiful one: the Russians love hard calculations to train their future mathematicians).
