"Different form" of Ceva's theorem? 
Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.


Now, the Art of Problem Solving site site reads one of the solutions as

Using a different form of Ceva's Theorem, we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$

I don't understand what this "different form" of Ceva's theorem is. How does one derive it?
 A: One way to prove it is to use Mass Points. We assign weights $a$ to vertex $A$, $b$ to $B$, $c$ to $C$. Then the weights of $F$ is $a+b$, of $D$ is $b+c$ of $E$ is $a+c$ and of $P$ is $a+b+c$.
Then we have that
$$\frac{AP}{PD} = \frac{b+c}{a} \implies \frac{AD}{PD} = \frac{a+b+c}{a} \implies \frac{PD}{AD} = \frac{a}{a+b+c}$$
Similarly:
$$\frac{PE}{BE} = \frac{b}{a+b+c} \quad \quad \frac{PF}{CF} = \frac{c}{a+b+c}$$
Adding this identities we have the wanted form.
Note that the method of mass points is another way of using Ceva's and Menelaus' Theorem. In fact all these identities can be proved using them, although it's a much more tedious process.

Here's a way to prove it purely by Ceva's and Menelaus Theorem.
Let $\frac{AF}{FB} = \frac ba$, $\frac{BD}{DC} = \frac cb$. Then by Ceva's Theorem we have that $\frac{CE}{EA} = \frac ac$. Now by Menelaus' Theorem on $B-P-E$ and $\triangle ADC$ we have that:
$$\frac{AP}{PD} = \frac{CB}{DB} \times \frac{EA}{CE} = \frac{b+c}{c} \times \frac ca = \frac{b+c}{a} \implies \frac{PD}{AD} = \frac{a}{a+b+c}$$
Similarly we get the other identities mentioned above. Hence the proof.
