Conjecture:

regardless of the position of P.

We begin by constructing two auxiliary lines parallel to segment AD through the vertices B and C. We then extend lines FC and BE to intersect these parallel lines at G and M respectively. The following similarities are apparent: triangle PDC ~ triangle GBC and triangle PDB ~ triangle MCB. Thus,

and

.

Solving the first equation for PD yields:

Substituting for PD in the second equation and simplifying:

.

The diagram also indicates that triangle BGF ~ triangle APF (Angles GFB and AFP are vertical angles and thus are congruent. Angles GBF and PAF are congruent because they are alternate interior angles formed by a transversal intersecting parallel lines. Therefore, triangle BFG ~ triangle APF by AA.) Triangles MCE and PAE are also similar. This implies that

and

.

Solving the equations for MC and BG respectively yields

and

.

We will take these expressions for BG and MC and substitute them into

.

Simplifying after the substitution results in

.

Mulitiplying both sides by (AE)/(CE):

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