Question 51
Let $P$ be a point inside a triangle $ABC$. The lines $AP$, $BP$ and $CP$ intersect the sides of $ABC$ in $A'$, $B'$ and $C'$ respectively. Posing
$$
x = \frac{AP}{PA'}, \quad y = \frac{BP}{PB'}, \quad z = \frac{CP}{PC'}
$$
prove that $xyz = x + y + z + 2$.
$$
x = \frac{AP}{PA'}, \quad y = \frac{BP}{PB'}, \quad z = \frac{CP}{PC'}
$$
prove that $xyz = x + y + z + 2$.