For any point inside a triangle

Question

Let $P$ be an interior point of the triangle $\triangle ABC$. Assume that $AP$, $BP$ and $CP$ meet the opposite sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Show that

$\frac{AF}{FB} +\frac{AE}{EC}= \frac{AP}{PD}$.

Answer 1

The statement is known as the Van-Aubel theorem. Here is a simple proof:

Apply Menelaus theorem for triangles $\triangle ACD$ and $\triangle ABD$. You get:

$$\frac {AP}{PD} * \frac {DB}{BC} * \frac {CE} {EA} = 1$$ $$\frac {AF}{FB} * \frac {BC}{CD} * \frac {DP} {AP} = 1$$

Now $$\frac {AE} {CE} = \frac {AP}{PD} * \frac {DB}{BC} $$ and $$\frac {AF}{FB} = \frac {CD}{BC} * \frac {AP} {PD} $$

Sum these equalities and you obtain what you wanted.