Two circles $\Gamma_1,\Gamma_2$ intersect at $A,B$. Through $B$ a straight line $\ell$ is drawn and $\ell\cap \Gamma_1=K,\ell\cap\Gamma_2=M\;(K,M\neq B)$. We are given $\ell_1\parallel AM$ is tangent to $\Gamma_1$ at $Q$. $QA\cap \Gamma_2=R\;(\neq A)$ and further $\ell_2$ is tangent to $\Gamma_2$ at $R$.
Prove that:
$\ell_2\parallel AK$
$\ell,\ell_1,\ell_2$ have a common point.
If $\ell$ cuts $\ell_1$ at $S,$ then it suffices to show that $RS$ is tangent to $\Gamma_2$ and parallel to $AK.$
Angle chasing using the tangency between $\ell_1,\Gamma_1$ and the cyclic quadrilateral $ARMB$ gives $\angle QBA=\angle SQR \equiv \angle SQA=\angle QAM=\angle SBR$ $\Longrightarrow$ $\angle ABR=\angle QBS$ and $QBRS$ is cyclic. Thus, from cyclic quadrilaterals $QBRS$ and $QBAK,$ we get $\angle QAK=\angle QBK \equiv \angle QBS=\angle QRS$ $\Longrightarrow$ $RS \parallel AK$ and $\angle ABR=\angle QBS=\angle QRS$ $\Longrightarrow$ $RS$ is tangent to $\Gamma_2,$ as desired.
Dear Mathlinkers,
another approach...
1. X the point of intersection of KM and the tangent l1
2. According to Reim’s theorem KQ // MR
3. According to a converse of the little Pappus theoreme applied to the hexagon XQKAMRX insert in the lines QAR and KBM, RX // AQ
4. by an simple angle chasing, we prove that RX is tangent to Gamma 2 at R.
Sincerely
Jean-Louis