If $P$ is an interior point of triangle $ABC$, prove $\frac{AF}{FB} + \frac{AE}{EC} = \frac{AP}{PD} $

Question

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

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

At first I tried to use the concept of similarity of triangles but that didn't help. Then, I thought about assigning some coordinates to the vertices (since that still confides the generality of the problem), but there are too many variables arising out of it.. so I had to bunk that idea as well.. I am just a high school student and do not have extensive knowledge when it comes to geometry..

How can I prove something with not one information given? Can anyone please help me out?

Thank you.

Answer 1

Let [] denote areas. Then,

\begin{align} \frac{AP}{PD} &= \frac{[ABE]}{[DBE]} = \frac{\frac{AE}{AC}[ABC] }{ \frac{BD}{BC}\frac{EC}{AC}[ABC] } = \frac{AE(BD+DC)}{EC \cdot BD } \\ &= \frac{AE}{EC } + \frac{AE}{EC } \frac{DC}{BD } = \frac{AE}{EC } + \frac{AF}{FB } \end{align}

where the Ceva’s theorem $\frac{AE}{EC } \frac{DC}{BD } \frac{BF}{FA }=1 $ is used in the last step.