Let $O$ is center of $\omega_2$ and $D,E$ are points of tangency of $AC$ and $BC$ Then $\angle A = \angle COB. \angle B= \angle COA$ $\angle ACO = \angle BCO \to AO=BO$ $AC^2+BC^2=(CD+AD)^2+(CD-AD)^2=2CD^2+2AD^2$ $\cos \angle COB+\cos \angle COA = \frac{CO^2+OB^2-CB^2}{2CO*OB}+\frac{CO^2+OA^2-AC^2}{2CO*OA} = \frac{2CO^2+2OA^2-2CD^2-2AD^2}{2CO*OA}= \frac{CD^2+DO^2+AD^2+DO^2-CD^2-AD^2}{CO*OA}=\frac{2OD^2}{CO*OA}=\frac{2OD^2 CA}{CO*OA*AC}=\frac{OD*4S_{COA}}{4S_{COA}OD}=1$

How is $AO=BO$ possible? My diagram says otherwise.

Hi: L ACO=L BCO is false, and O is a center of w, no is of w2, therefore AO=BO too is false. And I draw the problem, cos A+cos B= 1, is false, Sorry, my english is very bad. Thanks, Best Regards from Lima-Perú