I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
I had no idea how to find it nor where to start
Note that this picture are DRAWN ON SCALE
any explanation of how to find the area would be very appreciated
We can say, at least, that the option (C) is correct.
Let $p,q,r,s,t,u$ be the areas of white parts :
$p$ is the upper left area of $AB$, inside the circle whose diameter is $BC$.
$q$ is lower of $AB$, upper of $BC$, outside the circle whose diameter is $AC$.
$r$ is lower of $BC$, inside the circle whose diameter is $AB$.
$s$ is lower of $BC$, inside the circle whose diameter is $AC$.
$t$ is upper of $BC$, lower of $AC$, outside the circle whose diameter is $AB$.
$u$ is upper of $AC$, inside the circle whose diameter is $CB$.
Then, we have four equations :
For the circle whose diameter is $AB$ : $A_1+p=q+r+A_2\quad\tag1$
For the circle whose diameter is $AC$ : $u+A_3=A_2+t+s\quad\tag2$
For the circle whose diameter is $BC$ : $p+q+t+u+A_2=s+r+A_4\quad\tag3$
For the right triangle $ABC$ : $(A_1+p)+(u+A_3)=s+r+A_4\quad\tag4$
From $(1)$, we have $$A_1=A_2+q+r-p\tag5$$ From $(2)$, we have $$A_3=A_2+t+s-u\tag6$$ From $(3)$, we have $$A_4=A_2+p+q+t+u-s-r\tag7$$ From $(4)(5)(6)(7)$, we have $$(A_2+q+r-p)+p+u+(A_2+t+s-u)=s+r+(A_2+p+q+t+u-s-r),$$ i.e. $$A_2=p+u-s-r\tag8$$ From $(8)(1)(2)(3)$, we have $$A_1=q+u-s$$ $$A_3=p+t-r$$ $$A_4=2p+2u+q+t-2s-2r.$$
From these, we can have $$A_1+A_2+A_3=2p+2u+q+t-2s-2r=A_4.$$