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