Proving the following inequality in a triangle

Question

In a triangle the straight lines $AD$, $BE$, $CF$ are drawn through a point $P$ to meet $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively:

Prove that $$\frac{PD}{AD} + \frac{PE}{BE}+\frac{PF}{CF}=1$$ and $$\frac{AP}{AD}+ \frac{BP}{BE}+\frac{CP}{CF}=2$$

Answer 1

Let $G$ and $H$ on $AC$ so that $PG\parallel AB$ and $PH\parallel BC$. It follows that $$\frac{FP}{FC}=\frac{AG}{AC}, \frac{PD}{PA}=\frac{HC}{AC}.$$ Now we only need to show $$\frac{PE}{BE}=\frac{GH}{AC}.$$ But this follows from the fact that $$\frac{PE}{BE}=\frac{EG}{EA}=\frac{EH}{EC}=\frac{GE+EH}{AE+EC}.$$

Answer 2

$\dfrac{PD}{AD}$ is equal to $\dfrac{\mathrm{Area}\, BPC}{\mathrm{Area}\, ABC}$, and similarly for the other two ratios. Now just add them up. This gives you the first equation.

The second equation is simple arithmetic: $AP+PD=AD$, so $\dfrac{AP}{AD} = 1 - \dfrac{PD}{AD}$, and simliarly for the other two ratios. Now just add them up again.