Let's prove that $ C,A,S$ are collinear. Let $ \alpha = \frac12\angle{ACB}$. Then $ \angle{APQ} = \alpha$. Since $ AQCP$ is inscribed and $ SQ$ is tangent to $ k_1$, it follows that $ \alpha = \angle{APQ} = \angle{ACP} = \angle{AQP} = \angle{ACQ} = \angle{SQA}$. Since $ SB$ is tangent to $ k_2$, $ \angle{SBP} = \angle{PCB} = \alpha$. So $ \angle{SBP} = \angle{SBQ} = \angle{APQ}$, hence $ SB\|AP$. Let $ CA\cap SB = \{S'\}$. Since $ \angle{QCS'} = \alpha = \angle{QBS'}$, $ Q,C,B,S'$ are concyclic. Then $ \angle{S'CB} = 2\alpha = \angle{S'QB}$. But $ \angle{SQB} = \angle{SQA} + \angle{AQB} = 2\alpha$, hence $ S = S'$ and we're done. Similarly, $ C,B,T$ are collinear. Let $ y = \angle{PAB}$ and $ x = \angle{PBA}$. Since $ SB\|AP$ and $ TA\|BP$, $ \angle{ABS} = y$ and $ \angle{BAT} = x$. Apply now Sine Law in triangles $ ASB$ and $ ATB$ to obtain $ AS = \frac {AB\sin y}{\sin\angle{ASB}}$ and $ BT = \frac {AB\sin x}{\sin\angle{ATB}}$. From the above collinearities and parallelisms, one obtains $ \angle{ASB} = \angle{CAP}$ and $ \angle{ATB} = \angle{CBP}$, hence proving $ AS = BT$ is equivalent to proving $ \frac {\sin y}{\sin\angle{CAP}}\cdot\frac {\sin\angle{CBP}}{\sin x} = 1$, which is clear after applying Trigonometric Ceva for cevians $ CP,AP$ and $ BP$ in triangle $ ABC$.

http://www.math10.com/forumbg/viewtopic.php?t=5242

RotorM wrote: http://www.math10.com/forumbg/viewtopic.php?t=5242 Can you make '' quote '' on the Bulgarian text , then copy it and paste the solutions on this forum , translating only a few key words into english ? This would be great help. Babis

also discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=260084