A problem related to areas of triangles

Question

Consider a triangle ABC. there is a point P inside this triangle and BP and CP when extended meet AC and AB respectively at E and F.

Prove that $[BPC]^2>[BPF][CPE]$

Here [.] denotes area of the polygon

Answer 1

Let $[BPF]=u,[BPC]=v,[CPE]=w$

Join AP

Let $[AFP]=x,[AEP]=y$

From triangles AFC and BFC we have

$$x/(y+w)=FP/PC=u/v$$

Which gives

$$vx-uy=uw$$

Again from triangles AEB and CEB he have

$$vy-wx=uw$$

Solving these equations we get

$$x=uw(u+v)/(v^2-uw)$$

Now since $x$ is the area, $x>0$

Which means

$$v^2-uw>0$$

Or

$$v^2>uw$$

As required