$\left[\begin{array}{l}\text { Tangent Circles } \\ \text { [Isosceles, Inscribed, and Circumscribed Trapezoids }]\end{array}\right]$

Two circles touch externally at point $K$. One line is tangent to these circles at different points $A$ and $B$, and another line is tangent at different points $C$ and $D$. The common tangent to the circles passing through point $K$ intersects these lines at points $M$ and $N$. Find $MN$, if $AC=a, BD=b$.
$A B D C$ - is an isosceles trapezoid (or rectangle), and $M N$ is the midline.

## Solution

Let the lines $A B$ and $C D$ intersect at point $P$. The angles at the bases of the isosceles triangles $P A C$ and $P B D$ are equal, so $A C \| B D$. Therefore, $A B D C$ is an isosceles trapezoid. Since $M A = M K = M B$, $M$ is the midpoint of the lateral side $A B$. Similarly, $N$ is the midpoint of the lateral side $C D$. Therefore,

$$
M N = \frac{1}{2}(A C + B D) = \frac{a + b}{2}
$$

If $A B \| C D$, then $A B D C$ is a rectangle. In this case, $M N = A C = B D$.

## Answer

$\frac{a + b}{2}$.