In the adjoining figure, two Quadrants are touching at $$B$$. $$CE$$ is joined by a straight line, whose mid-point is $$F$$.

The measure of $$\angle CED$$ is ------.

Let the radius of the smaller quadrant be $$r$$, and the radius of the larger quadrant be $$R$$.

$$ \triangle CAB, \triangle EDB$$ are right angled isosceles triangles.

$$ CB = r\sqrt{2}, EB = R\sqrt{2}$$

Given, $$F$$ is the midpoint of $$CE$$.

Let $$ CF = CE = a$$

Applying pythagoras theroem in $$\triangle CBE$$

$$CB^2+BE^2 = CE^2 \implies 2r^2+2R^2 = (2a)^2 \ \text{or} \ r^2+R^2 = 2a^2$$ - Eq 1

Now, in triangle $$CBE, BF $$ is the median of the triangle.

Using Apollonius ' theorem, we can find the length of BF as 

 $$ 2(BF^2 + a^2) = 2r^2 + 2R^2 \implies BF^2 = R^2 +r^2 - a^2$$

From Eq 1, we can substitute the value of $$R^2+R^2$$, and we get

$$BF^2 = a^2 \implies BF = a$$.

Thus, we get that $$CF=BF=FE$$

Since $$BF = EF, \triangle BFE$$ is an isosceles triangle. $$\implies \angle FBE = \angle FEB$$

Now consider the following extension of the quadrant. 

$$ \angle BGE =\dfrac{ \angle BDE} {2} = 45\degree$$ ( Angle subtended at the centre is twice the angle subtended by the chord in the same sector)

$$ \angle BFE = 180\degree - 45\degree =135\degree$$ (Opposite angles in a cyclic quadrilateral are supplementary)

$$ \implies \angle FEB = \dfrac{180\degree-135\degree}{2} =22.5\degree$$ ( Angle sum in an isosceles triangle)

$$\angle CED = \angle FEB + \angle BED = 22.5\degree+45\degree = 67.5\degree$$