here we have another part of the problem

ABCD a rectangular
M a point

Prove : FBHM - GMED = ACKI

actually the two problems says that

x' = cos(a)*x + sin(a)*y
y' = cos(a)*y - sin(a)*x

formulas of rotation of coordinates

Here you need to help me. The previous one was simple by shear. Never thought of trigonometry.

https://www.cut-the-knot.org/Curriculum/Geometry/CirclesInRegularPolygon.shtml

in this scheme it is clear that we have two system of Cartesian coordinates.

in the two problems there is the same rectangular ABC - ACD and a point M.

in the second problem all we have to do is to prove that rectangulars with same color have equal areas

here is a hide

then

a1*a = b1*b + c1*c

a1 = b1 * b/a + c1 * c/a

a1 = b1*cos(C) + c1*sin(C)

but if you try to see this triangle ABC as the ACD in the second problem, then there is no problem.

i forgot,

if the point M is on the hypotenuse then

for two similar right triangles with sides a1,b1,c1 and a2,b2,c2

where a1 and a2 are the hypotenuses we have

b1*b2 + c1*c2 = a1*a2

c^2 = a^2 + b^2

c*c= a*a + b*b

k*c*c= k*a*a +k*b*b (k a positive number)

(k*c)*c = (k*a)*a + (k*b)*b

but k*a, k*b,k*c is a right triangle similar to a,b,c

https://www.cut-the-knot.org/pythagoras/index.shtml#13