Let $ABC$ be a triangle with circumcircle $\Gamma$. A line $d$ varies, parallel to $(BC)$, it intersects $(AB), (AC)$, and $\Gamma$ at $D, E, K$ and $L$ respectively. Let $\gamma_{1}$ (resp. $\gamma_{2}$) be the circle internally tangent to $\Gamma, [KD]$ and $[DB]$ (resp. $\Gamma, [LE]$ and $[EC]$).

What is the locus of the intersection of the internal tangents to the two inscribed circles?
We look at the involution in the quadrilateral $(A B),(A C),(B C),(D E)$, centered at $A$ which exchanges $D \leftrightarrow C, B \leftrightarrow E, A \leftrightarrow \infty$. After research, we find that it also exchanges $\gamma_{1}$ and $\gamma_{2}$. The axis of symmetry of our involution is the bisector of $\widehat{B A C}$, it exchanges $\left(A O_{1}\right)$ with $\left(A O_{2}\right)$, and by calling $T$ and $T^{\prime}$ the points of tangency of $\gamma_{1}$ and $\gamma_{2}$ to $(A B)$ and $(A C)$, we have $A O_{1} T \sim A O_{2} T^{\prime}$, hence $\frac{T O_{1}}{T^{\prime} O_{2}}=\frac{A O_{1}}{A O_{2}}$.

Next, let's study the negative homothety mapping $\gamma_{1}$ to $\gamma_{2}$, centered at the intersection $P$ of the internal tangents: we have $\frac{T O_{1}}{T^{\prime} O_{2}}=\frac{P O_{1}}{P O_{2}}$.

Therefore, $\frac{A O_{1}}{A O_{2}}=\frac{P O_{1}}{P O_{2}}$, and $P$ lies on the internal bisector of $\widehat{O_{1} A O_{2}}=\widehat{B A C}$.