Problem 1630

1. Define the incircles and their centers:
Let $\omega$ and $\omega'$ be the incircles of $\triangle ABC$ and $\triangle A'B'C'$, centered at $I$ and $I'$, respectively.

2. **Identify the intersection point $K$:**
Let $K = AC \cap A'C'$. By symmetry, $K$ lies on the line $II'$.

3. **Define the intersection point $T$:**
Let $T = AA' \cap II'$. The claim is that $T$ is the exsimilicenter of $\omega$ and $\omega'$.

4. Setup additional intersection points:
Define the following points:
- $P = A'C' \cap BC$
- $P' = AC \cap B'C'$
- $Q = A'C' \cap BA$
- $Q' = AC \cap B'A'$

5. Apply Pascal's theorem:
By Pascal's theorem on the hexagon $A'C'B'BCA$, we get that the lines $AA'$, $BB'$, and $P'$ are concurrent. Similarly, lines $CC'$, $BB'$, and $QQ'$ are concurrent.

6. Apply DDIT (Desargues' Dual Involution Theorem):
We will apply DDIT four times to establish involutions at $T$:
- On $AA'PP'$ at $T$, we get an involution $\Phi_1$ swapping $\{TA, TP\}$, $\{TA', TP'\}$, $\{TK, TB\}$.
- On $CC'QQ'$ at $T$, we get an involution $\Phi_2$ swapping $\{TC, TQ\}$, $\{TC', TQ'\}$, $\{TK, TB\}$.
- On $CBQK$ with incircle $\omega$ at $T$, we get an involution $\Phi_3$ swapping $\{TC, TQ\}$, $\{TA, TP\}$, $\{TK, TB\}$. It also swaps the two tangents from $T$ to $\omega$.
- On $C'B'Y'K$ with incircle $\omega'$ at $T$, we get an involution $\Phi_4$ swapping $\{TC', TQ'\}$, $\{TA', TP'\}$, $\{TK, TB\}$. It also swaps the two tangents from $T$ to $\omega'$.

7. Compare the involutions:
Observe that $\Phi_1$ shares two reciprocal pairs with both $\Phi_3$ and $\Phi_4$, and so does $\Phi_2$. Thus, we deduce that $\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4 = \Phi$.

8. Conclude the tangents must coincide:
If the tangents from $T$ to $\omega$ and $\omega'$ were different, then $\Phi$ would be the reflection across $TK$, which is absurd because it swaps $\{TK, TB\}$. Thus, the tangents must coincide, and $T$ is indeed the exsimilicenter.

9. Apply Monge's theorem:
Consider a circle $\Omega$ tangent to segments $BA$, $BC$, and $B'A'$. By Monge's theorem, the exsimilicenter of $\Omega$ and $\omega'$ must lie on $BT$. However, it must also lie on $B'A'$, so it is indeed $B'$, and thus $B'C'$ is also tangent to $\Omega$.

10. Conclusion:
Since $XBYB'$ is a convex quadrilateral and it has an incircle, the proof is complete.

$\blacksquare$