Circles with diameters equal to corresponding sides of the triangle

Let
a = length of side AB
b = length of side BC and
c = length of side AC
 

$A_{O1} = \frac{1}{4}\pi a^2$   →   Area of circle with diameter AB

$A_{O2} = \frac{1}{4}\pi b^2$   →   Area of circle with diameter BC

$A_{O3} = \frac{1}{4}\pi c^2$   →   Area of circle with diameter AC
 

$A_4 = A_{O3} - A_{O1} + A_1 - A_{O2} + A_2 + A_3$

$A_4 = \frac{1}{4}\pi c^2 - \frac{1}{4}\pi a^2 + A_1 - \frac{1}{4}\pi b^2 + A_2 + A_3$

$A_4 = (\frac{1}{4}\pi c^2 - \frac{1}{4}\pi a^2 - \frac{1}{4}\pi b^2) + A_1 + A_2 + A_3$

$A_4 = \frac{1}{4}\pi (c^2 - a^2 - b^2) + (A_1 + A_2 + A_3)$

$A_4 = \frac{1}{4}\pi \, [ \, c^2 - (a^2 + b^2) \, ] + (A_1 + A_2 + A_3)$
 

From Pythagorean Theorem
${AB}^2 + {BC}^2 = {AC}^2$

$a^2 + b^2 = c^2$

 

Thus,
$A_4 = \frac{1}{4}\pi \, [ \, c^2 - c^2 \, ] + (A_1 + A_2 + A_3)$

$A_4 = A_1 + A_2 + A_3$       (okay!)