How do I find the diameters of the circles in this geometry puzzle? My family and I like to do a daily quiz but this particular question has had us baffled for weeks. Please help. We only have basic mathematical knowledge.
 
 A: Let $d$ be the length of the segment $DC$, and $r$ the radius of the larger circle.  Thanks to the geometric mean theorem of elementary geometry we can write:
$$ \begin{align}
d \cdot r &= (r-5)^2 \\
d+r &= 2r - 9 \enspace.
\end{align}$$
This simplifies to
$$ \begin{align}
d \cdot r &= r^2 - 10r + 25 \\
d &= r - 9 \enspace.
\end{align}$$
Therefore the radius of the larger circle is $25$ cm and the radius of the smaller circle is $(2\cdot 25 - 9) / 2 = 41/2 = 20.5$ cm.
A: The diagram you need to draw, with $r$ as the radius of the larger circle (giving $r{-}\frac 92$ as the radius of the smaller) is: 

where $G$ is the centre of the smaller circle. From here you should be able to use Pythagoras to solve.

Since suitable time has now elapsed, the completion to a solution should look something like:
$$\require{cancel}\begin{align}
\left( r-\frac 92 \right)^2 &= (r-5)^2+\left( \frac 92 \right)^2 \\
r^2 - 9r +\left( \frac 92 \right)^2 &= r^2 - 10r + 25 +\left( \frac 92 \right)^2 \\
\cancel{r^2} - 9r +\cancel{\left( \frac 92 \right)^2} &= \cancel{r^2} - 10r + 25 +\cancel{\left( \frac 92 \right)^2}  \\
10r-9r&=25\\
r &= 25
\end{align}$$
So the diameter of the large circle is $2\cdot 25 = \fbox{50}$ and of the smaller circle $50-9 =\fbox{41}$ 
A: $\triangle DCE$, $\triangle ECA$ and $\triangle AED$ are right-angled triangles. We therefore have
$$ \begin{align}
    AD^2 &= AE^2 + DE^2 \\
         &= AC^2 + CE^2 + EC^2 + CD^2 \\
         &= AC^2 + (AC - 5)^2 + (AC - 5)^2 + (AC - 9)^2
\end{align}$$
But we also have $AD = 2AC - 9$, so we can solve for $AD$:
$$ \begin{align}
    (2AC - 9)^2 &= AC^2 + (AC - 5)^2 + (AC - 5)^2 + (AC - 9)^2 \\
    4AC^2 - 36 AC + 81 &= AC^2 + 2AC^2 - 20AC + 50 + AC^2 - 18AC + 81 \\
    2AC &= 50 \\
    AD &= 41
\end{align}$$
