 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): Return