27. Denote by $\alpha_{1}, \alpha_{2}, \alpha_{3}$ the angles of $\triangle A_{1} A_{2} A_{3}$ at vertices $A_{1}, A_{2}, A_{3}$ respectively. Let $T_{1}, T_{2}, T_{3}$ be the points symmetric to $L_{1}, L_{2}, L_{3}$ with respect to $A_{1} I, A_{2} I$, and $A_{3} I$ respectively. We claim that $T_{1} T_{2} T_{3}$ is the desired triangle. Denote by $S_{1}$ and $R_{1}$ the points symmetric to $K_{1}$ and $K_{3}$ with respect to $L_{1} L_{3}$. It is enough to show that $T_{1}$ and $T_{3}$ lie on the line $R_{1} S_{1}$. To prove this, we shall prove that $\angle K_{1} S_{1} T_{1}=\angle K^{\prime} K_{1} S_{1}$ for a point $K^{\prime}$ on the line $K_{1} K_{3}$ such that $K_{3}$ and $K^{\prime}$ lie on different sides of $K_{1}$. We show first that $S_{1} \in A_{1} I$. Let $X$ be the point of intersection of lines $A_{1} I$ and $L_{1} L_{3}$. We see from the triangle $A_{1} L_{3} X$ that $\angle L_{1} X I = \alpha_{3} / 2 = \angle L_{1} A_{3} I$, which implies that $L_{1} X A_{3} I$ is cyclic. We now have $\angle A_{1} X A_{3} = 90^{\circ} = \angle A_{1} K_{1} A_{3}$; hence $A_{1} K_{1} X A_{3}$ is also cyclic. It follows that $\angle K_{1} X I = \angle K_{1} A_{3} A_{1} = \alpha_{3} = 2 \angle L_{1} X I$; hence $X_{1} L_{1}$ bisects the angle $K_{1} X_{1} I$. Hence $S_{1} \in X I$ as claimed. Now we have $\angle K_{1} S_{1} T_{1} = \angle K_{1} S_{1} L_{1} + 2 \angle L_{1} S_{1} X = \angle S_{1} K_{1} L_{1} + 2 \angle L_{1} K_{1} X$. It remains to prove that $K_{1} X$ bisects $\angle A_{3} K_{1} K^{\prime}$. From the cyclic quadrilateral $A_{1} K_{1} X A_{3}$ we see that $\angle X K_{1} A_{3} = \alpha_{1} / 2$. Since $A_{1} K_{3} K_{1} A_{3}$ is cyclic, we also have $\angle K^{\prime} K_{1} A_{3} = \alpha_{1} = 2 \angle X K_{1} A_{3}$, which proves the claim.