Proving that a smaller inscribed, tangent circle to another circle makes an angle bisector of the given angle, joining the two tangent points

Question

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:

1. 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 $K$.
2. 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.
3. 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.

What am I missing? How to prove it?

Answer 1

Let $I$ be the centre of the inner circle. Join $IM$.

Since $AB$ is a tangent of the inner circle, $IM\perp AB$. Since $EF\parallel AB$ from your progress, also $IM\perp EF$.

So $M$ is the midpoint of arc $\overset{\Large\frown}{EMF}$. At point $K$, the angles that arcs $\overset{\Large\frown}{EM}$ and $\overset{\Large\frown}{MF}$ subtend are equal:

$$\angle EKM = \angle MKF$$

Answer 2

Here is a proof without angles.

First of all, the large circle is the image of the small one by the homothety (= enlargment) with center $K$ and ratio $r>1$ (equal to the ratio of their radii).

Let us express in two different ways the power of points $A$, resp. $B$, with respect to the small circle :

$$\begin{cases}AE \times AK &=& AM^2\\BF \times BK &=& BM^2\end{cases}\implies \frac{AE \times AK}{BF \times BK}=\frac{AM^2}{BM^2}\tag{1}$$

Besides :

$$AK=r EK = r(AK-AE) \ \iff \ r AE = (r-1) AK \ \iff $$

$$AE=\frac{r-1}{r}AK\tag{2}$$

For the same reason :

$$BF=\frac{r-1}{r}BK\tag{3}$$

Plugging (2) and (3) into (1) gives :

$$\left(\frac{AK}{BK}\right)^2=\left(\frac{AM}{BM}\right)^2 \ \implies \ \frac{AK}{BK}=\frac{AM}{BM}$$

We can conclude using the characteristic property of an angle bissector in a triangle to divide the opposite side in the ratio of the two other sides.