It is really easy to misunderstand this problem TheDarkPrince wrote: Let each of circles $\alpha, \beta, \gamma$ touches two remaining circles externally, and all of them touche a circle $\Omega$ internally at points $A_1, B_1, C_1$ respectively. The common internal tangent to $\alpha$ and $\beta$ meets the arc $A_1B_1$ not containing $C_1$ at point $C_2$. Points $A_2$, $B_2$ are defined similarly. Prove that the lines $A_1A_2, B_1B_2, C_1C_2$ concur. Draw the tangential triangle $A_3B_3C_3$ of triangle $A_1B_1C_1$. Now observe that $\overline{A_2A_3}, \overline{B_2B_3}, \overline{C_2C_3}$ concur at the radical center of $\alpha, \beta, \gamma$ (they are the radical axes obviously). Then apply the following well-known lemma: Claim: Let $ABC$ be a triangle with intouch triangle $DEF$. Suppose $X,Y,Z$ lie on minor arcs $EF, FD, DE$ of the incircle, respectively with $AX, BY, CZ$ concurrent. Then $DX, EY, FZ$ are also concurrent. (Proof) Sine rule chase for trig-form Ceva Theorem. $\blacksquare$

anantmudgal09 wrote: It is really easy to misunderstand this problem I remember anantmudgal09 wrote: Claim: Let $ABC$ be a triangle with intouch triangle $DEF$. Suppose $X,Y,Z$ lie on minor arcs $EF, FD, DE$ of the incircle, respectively with $AX, BY, CZ$ concurrent. Then $DX, EY, FZ$ are also concurrent. This is the Steinbart-Rabinowitz theorem.

WizardMath wrote: anantmudgal09 wrote: It is really easy to misunderstand this problem I remember anantmudgal09 wrote: Claim: Let $ABC$ be a triangle with intouch triangle $DEF$. Suppose $X,Y,Z$ lie on minor arcs $EF, FD, DE$ of the incircle, respectively with $AX, BY, CZ$ concurrent. Then $DX, EY, FZ$ are also concurrent. This is the Steinbart-Rabinowitz theorem. @WizardMath how do you manage to remember so many theorems??

It is more about practice and experience, you don't have to manually "learn" all the theorems.

WizardMath wrote: anantmudgal09 wrote: Claim: Let $ABC$ be a triangle with intouch triangle $DEF$. Suppose $X,Y,Z$ lie on minor arcs $EF, FD, DE$ of the incircle, respectively with $AX, BY, CZ$ concurrent. Then $DX, EY, FZ$ are also concurrent. This is the Steinbart-Rabinowitz theorem. It is also related to Mumbai Region RMO 2014 Problem 6. See http://artofproblemsolving.com/community/c6h616655p5389042

Just for the sake of completeness of the theorem, we have the following Extended Steinbart-Rabinowitz theorem wrote: Let $ABC$ be a triangle with intouch triangle $DEF$. Suppose $X,Y,Z$ are points on the incircle. Then $AX, BY, CZ$ are concurrent iff either $DX, EY, FZ$ are concurrent or $DX \cap EF, EY \cap FD, FZ \cap DE$ are collinear.