Dear All My Friends,

Here is my solution for the problem.

Proof of lemma
AD//BE => Area(ABD) = Area(ADE)
BD//CE => Area(BCE) = Area(CDE)
Now we calculate areas of three triangles ADE, CDE, FDE. Denote ha, hc, hf as their altitude segments to common base DE respectively. These three altitude segments are parallel and all together in one side with respect to line A, B, C because D, E all together in one side with respect to line A, B, C. Since F is midpoint of AC and B is on segment AC, we have: hf = (ha + hc)/2. It means:
Area(DEF) = (Area(ADE) + Area(CDE))/2
or
Area(DEF) = (Area(ABD) + Area(BCE))/2
The proof is completed!
Note: if hb is altitude length of triangle BDE from B, using similarities:
ha/hb = AD/BE = BD/CE = hb/hc or hb^2 = hh*hc. It means:
Area(ADE)*Area(CDE) = Area(BDE)^2
We have two types of means of areas here but only can use arithmetic mean to prove Pythagorean theorem.

Proof of Pythagorean theorem
Suppose ABC is right triangle at C with side lengths a, b, c. It means AB is diameter of one circle, say (O) centered at O with radius c/2. Denote M as midpoint of arc ACB.
Construct a circle (Oa) centered at A passing C
Construct a circle (Ob) centered at B passing C
Line CM intersects (Oa) again at A'
Line CM intersects (Ob) again at B'

On the circle (O): arc(AM) = arc(BM) = 90 degree hence angle(ACA') = angle(BCB') = 45 degree. It means two triangles CAA', CBB' are right isosceles triangles at A and B respectively.
Now we use result in:
http://www.cut-the-knot.org/Curriculum/Geometry/TwoIntersectingCircles.shtml
with two circles (Oa), (Ob), line A'CB' and circle (O). From this, we have M is midpoint of A'B'.
Now we can use our lemma and calculate areas of triangles:

Area(AMB) = (Area(CAA') + Area(CBB'))/2
MO*AB/2 = (AA'*AC/2 + BB'*BC/2)/2
c/2*c/2 = (a*a/2 + b*b/2)/2
c^2 = a^2 + b^2
The proof is completed!

Best regards,
Bui Quang Tuan