Given a triangle ABC, construct triangle DEF of maximal area with A, B, C on the sides of DEF, and angles ∠ABD = ∠BCE = ∠CAF

Let α = ABD = BCE = CAF.
Angles BDA+DAB = π - α = DAC, hence ∠BDA = ∠BAC

Then results that D lies on the circle from where we see side AB under angle A.

This circle is tangent in A to AC.

Similarily with E and F : DEF is similar to ABC

Note : The three circles intersect in a Brocard point of ABC.

Finding the maximal area triangle DEF is then equivallent to finding maximal side DF,
for the triangle is similar to a fixed triangle : ABC.

We have then again the known problem of the maximal double chord.

DF is maximal when it is parallel to JK. Hence the construction.