Here is the problem:
A chord $AB$ is drawn in a circle. Another circle is tangent to this chord at the point $M$ and to the given circle at the point $K$. Prove that $KM$ is the bisector of the angle $AKB$
A figure to the problem:
*All the data that is not given in the problem's description is marked with blue or red.
On my own I couldn't solve it, neither could I with a hint from the textbook, which is:
Let the segments $AK$ and $BK$ intersect the circle of smaller radius at points $E$ and $F$, respectively. Prove that $EF \parallel AB$. To do this, draw a tangent to the two given circles, that they both will share, through the point K.
I have indeed proved that they are parallel, but don't see where to move next:
- After making all the additional drawings suggested by the hint, I've also put two points $L$ and $J$ on the tangent line that passes through the point $K$.
- By a theorem in my book(which I think is a commonly known fact), since the $LJ$ line is tangent to both circles: $\angle LKA = \frac{1}{2} \cup AK = \frac{1}{2} \cup EK \implies \angle KFE = \angle KBA \implies EF \parallel AB$. The same way it can be proven that $\angle JKB = \angle KEF = \angle KAB$, but it's not necessary, I guess.
- From this point on I've tried a variety of different approaches, but none of them brought me even a bit closer to the proof.