Goutham wrote: Let $k$ be the incircle and let $l$ be the circumcircle of the triangle $ABC$. Prove that for each point $A'$ of the circle $l$, there exists a triangle $(A'B'C')$, inscribed in the circle $l$ and circumscribed about the circle $k.$ Suppose, $ I $ is the center of $ k $. Now for any point $ A' $ on $ l $ draw tangents $ A'X,A'Y $ to $ k $ with $ X,Y $ lying on $ k $. Suppose, $ A'X,A'Y $ meet $ l $ at $ B',C' $. We have to prove that $ B'C' $ touches $ k $. The radius of $ k,l $ be $ r,R $. Note that under inversion wrt $ I $ with power $ r^2 $, $ l $ goes to a circle of radius $ \frac {r}{2} $. $ X,Y $ will remain fixed and $ A' $ will go to the midpoint $ M $ of $ XY $. Now, suppose, after inversion $ l $ goes to $ l' $ and $ N $ be the center of $ l' $. Now draw a triangle with $ l' $ as nine-point circle and $ k $ as circumcircle and two vertices being $ X,Y $. Such a triangle always exists since radius of $ l' $ is half of the radius of $ k $. Suppose, $ Z $ is the other vertex of that triangle. Midpoint of $ XZ,YZ $ are $ M_1,M_2 $. Note that the image of $ M_1,M_2 $ under the inversion are $ B',C' $ and $ Z $ is the tangency point of $ B'C' $ with $ k $.