$ABCD$ is a rectangle with area $S$ and $AC\cap BD=O$. The circumscribed circle of $\triangle ABO$ intersects for second time the line $AD$ at $M$, such that $\tan\measuredangle ABM=1$. Find the diagonal and the perimeter of the rectangle $ABCD$.
I think that we can say $\measuredangle ABM$ is an acute angle because it's one of the angles in the right-angled triangle $ABM$. Is this so? This means $\measuredangle ABM=45^\circ$ or $AB=AM$. So the length of the segment $DM$ is actually the sum of the sides of the rectangle $ABCD$.
It is worth noting that the centre of the circumscribed circle of $\triangle ABO$ lies on $BM$ because $ABM$ is also inscribed in that same circle and $\measuredangle BAM=90^\circ$. How does the given area $S$ come to play? Thank you!
Say $AD = a, AB = b$. Then $ab = S$
We also know $AM = AB = b$, given $\angle ABM = 45^\circ$
Using power of point $D$,
$OD \cdot BD = AD \cdot MD$
Or, $ \displaystyle \frac{BD^2}{2} = \frac{a^2+b^2}{2} = a \cdot (a+b)$
And we get, $b = (\sqrt2 + 1) \cdot a$
Now knowing relationship between $a$ and $b$ and also knowing that $ab = S$. Can you find $a$ and $b$ in terms of $S$? From there, you can find the diagonal and the perimeter.