58. National mathematical olympiad Final round

Denote by $d_c(X)$ the distance from the point $X$ to the line $AB$ and analogously for the lines $BC$ and $CA$. It is not difficult to see that the Sine version of the Ceva theorem implies that the equality $$\frac{d_b(A^{\prime })}{d_c(A^{\prime })}\cdot \frac{d_c(B^{\prime })}{d_a(B^{\prime })}\cdot \frac{d_a(C^{\prime })}{d_b(C^{\prime })}=1$$ is necessary and sufficient for the lines $A{A}^{\prime }$, $B{B}^{\prime }$ and $C{C}^{\prime }$ to be concurrent.

Note that $B_1{A}^{\prime }=A_1{B}^{\prime }$. Moreover, since the lines $CB$ and $CA$ are tangent to the incircle, we have $\angle B^{\prime }{A}_{1}B=\frac{1}{2}{B}^{\prime }{A}_1=\frac{1}{2}{A}^{\prime }{B}_1=\angle A^{\prime }{B}_{1}C$. Then $d_a(B^{\prime })=A_1{B}^{\prime }\sin \angle B^{\prime }{A}_{1}B=B_1{A}^{\prime }\sin \angle A^{\prime }{B}_{1}C=d_b(A^{\prime })$. We analogously obtain $d_b(C^{\prime })=d_c(B^{\prime })$ and $d_c(A^{\prime })=d_a(C^{\prime })$ which completes the proof.