Two circles $\Gamma_1,\Gamma_2$ have centers $O_1,O_2$. Let $\Gamma_1\cap\Gamma_2=A,B$, with $A\neq B$. An arbitrary line through $B$ intersects $\Gamma_1$ at $C$ and $\Gamma_2$ at $D$. The tangents to $\Gamma_1$ at $C$ and to $\Gamma_2$ at $D$ intersect at $M$. Let $N=AM\cap CD$. Let $l$ be a line through $N$ parallel to $CM$, and let $l\cap AC=K$. Prove that $BK$ is tangent to $\Gamma_2$.
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Here is some progress I have:
We are looking to prove $\angle O_2BP=90^{\circ}$, and since $\angle O_2DP=90^{\circ}$, if we could prove $BP=PD$, we would be done by congruent triangles. So we are looking to prove $\dfrac{\sin \angle BDP}{\sin \angle DBP}=1$. Let $AM\cap BK=l$. We have $\angle BDP=\angle QBN$. By the law of sines on $\triangle DNM,\triangle BNQ$, we have $\sin \angle BDP=\sin \angle NDM=\dfrac{NM}{DM}\sin \angle DNM$ and $\sin\angle QBN=\dfrac {NQ}{BQ}\sin\angle BNQ$. Dividing, the sines cancel (since they are supplementary), and we are left with $\dfrac{NM\times BQ}{DM\times NQ}$, so it remains to prove $\dfrac{NM}{DM}=\dfrac{NQ}{BQ}$.
I'm not sure what to do from here. We would be done if we could prove $\triangle NBQ\sim\triangle NDM$, but this would imply $\angle QNB=\angle MND=90^{\circ}$, but from drawing multiple diagrams, it looks like this is not always the case.
As always, any ideas are appreciated!
Let us temporarily remove some of the distracting lines first so that we can use the following diagram to show ADMC is cyclic.
That claim is true because $x + (y_1) + [z_1] = x + (y_2) + [z_2] = 180^0$.
Now we add back the lines KB, ANM, KN.
From the fact that “$z_3 = z_1 = z_ 2$”, we can say $KN // CM$.
From the fact that "$\angle 1 = \angle 2 = \angle 3 = \angle 4$”, we can conclude that KB is tangent to the circle ADB.