area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.

|AB| = a, P is midpoint of segment AB

Let $|AB| = a$, $|BC|=a$, $|CA|=a.$ By Pythag we know $|PC|=\sqrt{a^2 - (\frac{a}{2})^2}$ and therefore the area of the equilateral triangle $ABC$ is $\frac{a\left(\sqrt{a^2 - (\frac{a}{2})^2}\right)}{2}$. Now if $ABC$ is an equilateral triangle then the angle between $AB$ and $BC$ is $\pi/3$.
So we work out the area between the edge of the circle and the triangle $ABC$, there are 2 of these but they are equal. To do this we find the area of one of the segments of the circle $AB$, $BC$ which is $\frac{\pi a^2}{6}$ and then minus the area of the equilateral triangle, so we get:
\begin{equation}
X=\frac{\pi a^2}{6}-\frac{a\left(\sqrt{a^2 -(\frac{a}{2})^2}\right)}{2}
\end{equation}
So the total area in the $ABC$ area (the blue and the white circles combined) is:
\begin{equation}
\begin{aligned}
Y=& \frac{a\left(\sqrt{a^2 - (\frac{a}{2})^2}\right)}{2} +2X \\ =&
\frac{\pi a^2}{3} - \frac{a\left(\sqrt{a^2 -(\frac{a}{2})^2}\right)}{2}
\end{aligned}
\end{equation}
Now we have that area we just need to minus the area of the white circles inside it. The two semi-cicles on the line $AB$ are of same size and therefore we can just minus the area of the full circle equal to $\frac{\pi a^2}{16}$. Now we just need the area of the full circle at the top, call it $Z$. Then the blue area is equal to $Y - \frac{\pi a^2}{16} - Z$. Unfortunately at the moment I can't work out the area $Z$ but I thought i'd still share the rest of this answer and hopefully you can get it.
