Let $ABC$ be a triangle inscribed in the circle $K$ and consider a point $M$ on the arc $BC$ that do not contain $A$. The tangents from $M$ to the incircle of $ABC$ intersect the circle $K$ at the points $N$ and $P$. Prove that if $\angle BAC = \angle NMP$, then triangles $ABC$ and $MNP$ are congruent.

Valentin Vornicu

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1. Let $L$ be the incircle of $\triangle ABC$ with incenter $I$, and let $O$ be the center of the circumcircle $K$.
2. Given that $\angle BAC = \angle NMP$, we need to show that $\triangle ABC \cong \triangle MNP$.
3. Since $M$ is on the arc $BC$ that does not contain $A$, $M$ is the midpoint of the arc $BC$ in the circumcircle $K$. This implies that $OM = OA$.
4. The tangents from $M$ to the incircle $L$ intersect the circumcircle $K$ at points $N$ and $P$. By the given condition, $\angle BAC = \angle NMP$.
5. Consider the reflection of the entire configuration in the line $\overleftrightarrow{OI}$. Under this reflection:
   - The circumcircle $K$ and the incircle $L$ map to themselves because $O$ and $I$ are their centers, respectively.
   - Point $A$ maps to point $M$ because $OM = OA$.
   - Since $A$ maps to $M$, the points $B$ and $C$ must map to $N$ and $P$, respectively, to preserve the angles and tangents.
6. Therefore, the reflection implies that $\triangle ABC$ maps to $\triangle MNP$.
7. Since the reflection preserves distances and angles, $\triangle ABC \cong \triangle MNP$.

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