Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Suppose it's false, wlog $BC < CD$. Let $CK$ and $CL$ be the bisectors of the correspondent angles in triangles $\triangle ABC$ and $\triangle CDA$, and $B'$ and $D'$ be the points symmetric to the points $B$ and $D$ with respect to the lines $CK$ and $CL$ correspondingly (fig. 20). Then lines $KB$ and $KB'$ are symmetric with respect to the line $CK$, so $KB'$ is tangent to the inscribed circle of $\triangle ABC$, similarly $LD'$ is tangent to the inscribed circle of $\triangle CDA$.

Fig. 20

Denote by $d(X)$ the oriented distance from the point $X$ to the line $BD$, where $d(A) > 0$.
Note that $\angle DBC > \frac{1}{2}\angle BAD$, implying
$$\begin{aligned} \angle AKB' &= 180^\circ - 2\angle BKC \\ &= 2\angle ABC + \angle ACB - 180^\circ \\ &> 2\angle ABD + \angle BAD + \angle ADB \\ &\ge -180^\circ = \angle ABD, \end{aligned}$$
so $d(B') < d(K)$. Similarly, $\angle ALD' < \angle ADB$, so $d(D') > d(L)$. Also note that
$$\frac{AK}{KB} = \frac{AC}{CB} > \frac{AC}{CD} = \frac{AL}{LD} \Rightarrow d(K) < d(L).$$
Consider the line $k \parallel BD$, with respect to which the points $K$ and $L$ lie on different sides. Then the inscribed circle of $\triangle ABC$ lies on the same side from the line $k$, as the segment $BD$ (as this circle lies in the quadrilateral $BCB'K$, which lies on this side with respect to $k$) and the inscribed circle of $\triangle CDA$ has a point, lying from the different side with respect to $k$ (as it's tangent to the segment $LD'$, which lies from the different side with respect to $k$). So these inscribed circles can't have a common tangent, parallel to $BD$, which lies on the same side from $BD$, as point $A$, this contradiction completes the proof.

Also note that when $C$ is the midpoint of arc $BD$, we have $CB' = CB = CD = CD'$, so $B' = D'$ is the incenter of $\angle DAB$ (fig. 21). Furthermore, $\angle AKB' = 180^\circ - 2\angle BKC = 2\angle ABC + \angle ACB - 180^\circ = 2\angle ABD + \angle BAD + \angle ADB - 180^\circ = \angle ABD$, so $KB' \parallel BD$. Similarly $LD' \parallel BD$, so the line $KL$ is the common tangent.

Fig. 21