NOTE: Please WAIT while the applet below loads.
The beauty of this problem is that it can be proved in many different ways ranging from different synthetic approaches involving similarity, Pascal's theorem, enlargement, power of a point, parallelograms, etc. to computational ones involving areal or normal coordinates, trigonometry, etc. The proof below using a homothetic transformation (enlargement) is particularly elegant.
Homothetic proof
Proof: Let X and Y be the midpoints of sides AB and AC, respectively, and Z the intersection of PX and QY as shown in Figure 1. Then PX and QY are medians of similar triangles ABP and CAQ. Thus, ∠BXP = ∠QYA and therefore quadrilateral AXZY is cyclic (exterior angle equals opposite interior angle). Now the homothety with centre A and ratio 2 maps AXZY to ABLC, and completes the proof.
The result about the equality of the products of the opposite sides of the cyclic quadrilateral ABLC, namely AB*CL = BL*AC = ½AL*BC, also follows from the similarities in the figure, and an application of Ptolemy's theorem.
More Proofs: Additional proofs are available as downloadable PDF's at two simple proofs or more proofs.
Created by Michael de Villiers, 16 July 2014; modified & updated to WebSketchpad on 4 July 2021