Euclid's Riddle

First we observe that $\angle ABF=\angle ACF=90^\circ$ since the angle subtended by a semi-circle is always $90^\circ.$
Let the radius be $r, AB=x.$
Since $AB+AC=BE+CD\Rightarrow x+8=7+CD\Rightarrow CD=x+1.$
We get $BF=\sqrt{4r^2-x^2}$ and $CF=\sqrt{4r^2-64}$ by Pythagoras Theorem .
We carefully observe that $\triangle ABF\sim \triangle FBE.$
Using similarity properties , $\dfrac{EB}{FB}=\dfrac{FB}{AB}\Rightarrow \dfrac{7}{\sqrt{4r^2-x^2}}=\dfrac{\sqrt{4r^2-x^2}}{x}\Rightarrow 4r^2=7x+x^2. - - - - 1.$
Again we see that $\triangle ACF\sim\triangle FCD.$
Using similarity properties , $\dfrac{CD}{FC}=\dfrac{FC}{AC}\Rightarrow \dfrac{x+1}{\sqrt{4r^2-64}}=\dfrac{\sqrt{4r^2-64}}{8}\Rightarrow 4r^2=8x+72 - - - - 2.$
Since $1$ and $2$ are equal, $7x+x^2=8x+72\Rightarrow x=9.$
$\therefore 4r^2=7x+x^2\Rightarrow 4r^2=144\Rightarrow r=\boxed{6}.$

According to the Euclid's chord theorem, considering $\triangle AEF$ , $EF^2 = AE \cdot BE$ .

Since $\triangle AEF$ is a right triangle, we can also apply Pythagorean theorem: $AE^2 = EF^2 + AF^2$ .

Hence, $AE^2 = (AE)(AB + BE) = EF^2 + AF^2 = AE \cdot BE + AF^2$ .

Subtracting the common terms: $AE \cdot AB = AF^2$ .

In other words, for a right triangle with one of the adjacent sides as diameter of the superimposed circle, the product of the internal chord and the hypotenuse equals to the square of the diameter.

That means, $AF^2 = AB\cdot AE = AC\cdot AD$ .

We know that the sum of the red chords equal to that of the blue lines, so we can set up the equation by letting $AB = x$ .

Hence, $x + 8 = 7 + CD$ . $CD = x +1$ .

Then $x(x+7) = 8(8 + (x+1))$

$x^2 + 7x = 8x +72$

$x^2 -x +72 = 0$

$(x-9)(x+8) = 0$

Thus, $x = AB = 9$ , and $CD = 10$ .

Plugging in the values to calculate the diameter:

$AF^2 = AB\cdot AE = 9(7+9) = 144$

$AF^2 = AC\cdot AD = 8(8+10) = 144$

The diameter $AF = 12$ . Therefore, the radius of the circle is $\boxed{6}$ .

(Euclid should be proud now.)

Put AB AS x, BE as 7 Ac as 8 CD as b EF as p and FD as q. then x+8=7+b (1) (x+7)^2-p^2=(2r)^2 r is radius (2) (8+b)^2-q^2=(2r)^2 (3) Also intersecting chords theorem for AE and EF gives 7(7+x)=p^2 and (8+b)b= (q)^2 (4and 5)

Solving these 5 equations radius was found as 6