jmc

Let $a=OA=OB$ and $b=OC=OD.$ Then $a^2-b^2=O{F}^2=36.$



In general, if a right triangle has legs $x$ and $y,$ and hypotenuse $z,$ then its inradius is given by $$\frac{x+y-z}{2}.$$Hence, the diameter of the incircle of triangle $OCF$ is $$OC+OF-CF=2.$$Then $b+6-a=2,$ so $a-b=4.$

By difference of squares on the equation $a^2-b^2=36,$ $(a+b)(a-b)=36,$ so $$a+b=\frac{36}{a-b}=9.$$With the equation $a-b=4,$ we can solve to get $a=\frac{13}{2}$ and $b=\frac{5}{2}.$

Then $AB=13$ and $CD=5,$ so $(AB)(CD)=\boxed{65}.$