Prove that (AF)(BD)(CE)=(BF)(CD)(AE), and :

Given AD, BE, and CF are concurrent at point P, construct a line parallel to BC, through A and extend CF and BE to intersect this line at G and H respectively.

From the above diagram, we can establish angle AGP congruent to angle DCP, and angle GAP is congruent to angle CDP because alternate interior angles are congruent. So triangle GAP is similar to triangle CDP by angle-angle. Therefore, because corresponding sides of similar triangles are proportional.

From the above diagram, we can establish angle AHP congruent to angle DBP, and angle HAP is congruent to angle BDP because alternate interior angles are congruent. So triangle HAP is similar to triangle BDP by angle-angle. Therefore, because corresponding sides of similar triangles are proportional.

By substitution, , or

From the above diagram, we can establish angle AHE congruent to angle CBE, and angle HAE is congruent to angle BCE because alternate interior angles are congruent. So triangle AEH is similar to triangle CEB by angle-angle. Therefore, because corresponding sides of similar triangles are proportional.

From the above diagram, we can establish angle GAF congruent to angle CBF, and angle AGF is congruent to angle BCF because alternate interior angles are congruent. So triangle GAF is similar to triangle CBF by angle-angle. Therefore, because corresponding sides of similar triangles are proportional.

So by the multiplicative property of equality. Furthermore, by the property of multiplicative inverses. Therefore by the transitive property, or