The message is related Maximum Area Property of Equilateral Triangles:
https://www.cut-the-knot.org/triangle/InscribedEquilateral.shtml

I think after proofing that isosceles triangle has area bigger than any triangle with common base, we can have another proof as following by infinite continue of changing apex of isosceles triangles.

We start with any triangle ABC inscribed in given circumcircle and infinite continue of making the triangle isosceles with apex: A, B, C, A, B, C...

At least one of three angles A, B, C must be less or equal 60 (degree) so we can suppose A <= 60.

If we make the triangle isosceles at one angle (vertex X) so these angle X is not changed, only two other changed and equal (180 - X)/2

The angles we can get after each time as following:
A, B, C
A, (180-A)/2, (180-A)/2
(180-(180-A)/2)/2, (180-A)/2, (180-(180-A)/2)/2
...

So, we get a sequence a(n) of angles:
A, (180-A)/2, (180-(180-A)/2)/2,...

here: a(0)=A, a(n+1) = (180-a(n))/2 with n=1,2,3...

If 0<=X<=60 then X<=(180-X)/2 and (180-X)/2>=60
If X>=60 then X>=(180-X)/2 and (180-X)/2<=60
If 0<=X<=60 then (180-(180-X)/2)/2>=X and (180-(180-X)/2)/2 <=60
If X>=60 then (180-(180-X)/2)/2<=X and (180-(180-X)/2)/2 >=60

Since we choose A<=60 so a(n) can be devided into two sub monotone sequences:
a(0)<=a(2)<=a(4)<=a(6)<=... <=60
a(1)>=a(3)>=a(5)>=a(7)>=... >=60

Therefore these two subsequences have limits. Since both have common form f=(180-f)/2 so they have common limit 60. It means we can get, at the end, equilateral triangle and equilateral triangle has maximum area

Best regards,
Bui Quang Tuan