angleeuclidean-geometrygeometryplane-geometry
For reference:
(figure without scale)
My progress:
Circle centers are collinear.
I think that the straight line joining the centers passes through the point of tangency with the side of the triangle. Therefore, the triangles formed will be rectangles and $\measuredangle a$ and $\measuredangle b$ will be $90^\circ$. It remains to demonstrate that $\measuredangle c = 45^\circ$.
Update:
Draw $AJ \parallel CE$ such that $AJ = CE$. Use symmetry about diagonal $AC$.
$\triangle ACE$ is a well known right triangle with perpendicular sides in the ratio $1:2$. The angles are $26.5^\circ$ and $63.5^\circ$.
So, $\angle EAD = \angle 63.5^\circ - 45^\circ = 18.5^\circ$
Here is another observation. Points $A, B, J, C, D$ and $E$ are all concyclic with center at $O$.
Here is a complete diagram with $9$ squares.
$\angle BAC + \angle FAE = 65^0$
$\angle PAQ = \frac{\angle POQ}{2} = 90^\circ - \frac{x}{2}$
$\angle BAF + \angle BDF = 180^\circ$
$\implies 65^\circ + 90^\circ - \frac{x}{2} + x = 180^\circ$
$\implies x = 50^\circ$
Best Answer
Yes, you are correct, $\measuredangle b$ is $90^\circ$, but not $\measuredangle a$.
Hint: Consider the pentagon (sum of interior angles=$540^\circ$). Can you take it from here?