62nd 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 $△ABC$ and $△CDA$, and $B^{\prime }$ and $D^{\prime }$ 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 $K{B}^{\prime }$ are symmetric with respect to the line $CK$, so $K{B}^{\prime }$ is tangent to the inscribed circle of $△ABC$, similarly $L{D}^{\prime }$ is tangent to the inscribed circle of $△CDA$.

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 $$\begin{array}{rl}\angle AK{B}^{\prime }& =180^{\circ }-2\angle BKC\\ & =2\angle ABC+\angle ACB-180^{\circ }\\ & >2\angle ABD+\angle BAD+\angle ADB\\ & \ge -180^{\circ }=\angle ABD,\end{array}$$ so $d(B^{\prime })<d(K)$. Similarly, $\angle AL{D}^{\prime }<\angle ADB$, so $d(D^{\prime })>d(L)$. 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 $△ABC$ lies on the same side from the line $k$, as the segment $BD$ (as this circle lies in the quadrilateral $BC{B}^{\prime }K$, which lies on this side with respect to $k$) and the inscribed circle of $△CDA$ has a point, lying from the different side with respect to $k$ (as it's tangent to the segment $L{D}^{\prime }$, 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 $C{B}^{\prime }=CB=CD=C{D}^{\prime }$, so $B^{\prime }=D^{\prime }$ is the incenter of $\angle DAB$ (fig. 21). Furthermore, $\angle AK{B}^{\prime }=180^{\circ }-2\angle BKC=2\angle ABC+\angle ACB-180^{\circ }=2\angle ABD+\angle BAD+\angle ADB-180^{\circ }=\angle ABD$, so $K{B}^{\prime }\parallel BD$. Similarly $L{D}^{\prime }\parallel BD$, so the line $KL$ is the common tangent.

Fig. 21