Find the ratio $AB:CD$ in the figure below

Question

From the graph, calculate $\dfrac{AB}{CD}$. (Answer :$1$)

I didn't get much.

$AMBO$ is cyclic

$OCPK$ is cyclic

I tried to find a similarity of triangles but I had no success.

Answer 1

In your drawing (second one) there are two quadrilaterals MABO and CODP, each one inscribed in a circle of diameter OM=OP.

As a cord, AB opposes $\angle AMB=\angle AOB=\angle COD$, the last angle opposed to cord CD

Since AB and CD oppose the same measure angle with vertex on same radius circles, then $\frac {AB}{CD}=1$.

Answer 2

This is what I could costruct:

In circle passing O, F, C and D we have:

$\widehat{EFC}=\widehat{COE}=\widehat{AOB}$

Also:

$\widehat{BAO}=\widehat{OEF}=\widehat{CEF}$

$EF=AO$

So right angled triangles AOB and ECF are congruent for ASA , so $EC=AB$.