blues

the above is hippocrates theorem

the below is proof

using The Pythagorean Theorem we can prove that : the area of the semicircle on AC = the sum of areas of the semicircles on AB and BC. ( By the way, this can be considered as an other extension of the Pythagorean Theorem).

Therefore ; the area of the semicircle on AC = the green area + the white area.

the sum of areas of the semicircles on AB and BC = the blue area + the white area

so far , the blue area = the green area = 24

AC is the diameter of the semi-circle formed by arcs AB and BC, given the theorem that any triangle with a base as the diameter of a circle will form a right angled triangle.

Area of the white portion = Area of the semi-circle mentioned above - Area of triangle ABC

= $\pi$ x $(AC/2)^{2}$ - (0.5 x AB x BC)

Given $AC^{2}$ = $BC^{2}$ + $AB^{2}$

= $\pi$ x [ $(BC/2)^{2} + (AB/2)^{2}$ ] - (0.5 x AB x BC) ... (1)

Area of the semi-circles formed by the two lunes = $\pi$ x [ $(BC/2)^{2} + (AB/2)^{2}$ ] ... (2)

Required area of the two blue lines = (2) - (1) = (0.5 x AB x BC) = 0.5 x 6 x 8 = 24 square units

The area of the two blue lunes is equal to the area of the green triangle.