Point inside a triangle 
Let $\triangle ABC$ be an acute triangle and let $P$ be a point inside of it. $PD$, $PE$ and $PF$ are respectively parallel to sides $AC$, $BC$ and $AB$. Prove that $$\frac{|BE|}{|AB|}+\frac{|CD|}{|BC|}+\frac{|AF|}{|AC|}=1$$

Could someone give me a hint? I already extended $PD$, $PE$ and $PF$ and wrote down all the similar triangles and the ratio's, but that doesn't seem to help..

 A: Thales is your friend.
You have
$\dfrac{|BE|}{|AB|}=\dfrac{|PD|}{|AC|}$ (Thales)
$\dfrac{|BE|}{|AB|}+\dfrac{|CD|}{|BC|}+\dfrac{|AF|}{|AC|}=\dfrac{|PD|+|AF|}{|AC|}+\dfrac{|CD|}{|BC|}$
But $|PD|+|AF|=|D'D|$, with $D'$ the projection of $P$ along the direction $(AC)$ onto $(AB)$
And $\dfrac{|DD'|}{|AC|}=\dfrac{|BD|}{|BC|}$ (Thales)
Thus 
$\dfrac{|BE|}{|AB|}+\dfrac{|CD|}{|BC|}+\dfrac{|AF|}{|AC|}=\dfrac{|PD|+|AF|}{|AC|}+\dfrac{|CD|}{|BC|}=\dfrac{|DD'|}{|AC|}+\dfrac{|CD|}{|BC|}=\dfrac{|CD|}{|BC|}+\dfrac{|BD|}{|BC|}=1$
A: Vector solution — mechanical solution.
Let's $\overrightarrow{AB} = \vec a$, $\overrightarrow{AC} = \vec b$; since $E$, $F$ and $D$ lies on $AB$, $AC$ and $BC$, we have $\overrightarrow{AE}=\alpha \vec a$, $\overrightarrow{AF}=\beta \vec b$ and $\overrightarrow{BD}=\gamma\overrightarrow{BC}=\gamma (\vec b-\vec a)$. We should prove that
$$
(1-\alpha) + (1-\beta) + \gamma = 1 \Longrightarrow \fbox{$1 + \beta = \alpha + \gamma$}.
$$
Ok, let's start. We know that $EP\parallel BC$, $FP\parallel AB$, $PD\parallel AC$. Hence $\overrightarrow{EP}=\delta (\vec b - \vec a)$, $\overrightarrow{FP}=\varepsilon \vec a$, $\overrightarrow{PD}=\zeta \vec b$. Now we express $\overrightarrow{AP}$ by two ways:
$$
\overrightarrow{AP} = \overrightarrow{AE} + \overrightarrow{EP} = \alpha\vec a + \delta (\vec b - \vec a) = (\alpha-\delta)\vec a + \delta \vec b,
$$
$$
\overrightarrow{AP} = \overrightarrow{AF} + \overrightarrow{FP} = \beta \vec b + \varepsilon \vec a.
$$
We conclude that
$$
\alpha - \delta = \varepsilon,\\
\delta = \beta.
$$
In a same manner we express $\overrightarrow{AD}$:
$$
\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \vec a + \gamma(\vec b - \vec a)
$$
$$
\overrightarrow{AD} = \overrightarrow{AP} + \overrightarrow{PD},
$$
and
$$
\left\{\begin{aligned}
\alpha-\delta=1-\gamma\\
\alpha + \zeta = \gamma,\\
\varepsilon = 1-\gamma,\\\beta + \zeta = \gamma
\end{aligned}\right.
$$
(from two expressions for $\overrightarrow{AP}$). From first equation $\alpha - \delta = 1 - \gamma = \alpha - \beta$, or $\alpha + \gamma = 1 + \beta$. That's all.
