Let $ABC$ be a triangle with $AB<AC<BC$. Let the incircle of $ABC$ be tangent to $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. Let ${\omega }_A$ be the circle through $A$ tangent to $BC$ at $D$, let ${\omega }_B$ be the circle through $B$ tangent to $CA$ at $E$, and let ${\omega }_C$ be the circle through $C$ tangent to $AB$ at $F$.

Let ${\omega }_B$ and ${\omega }_C$ meet at $P\ne B$ and $Q\ne C$ (so that $P$ is closer to $B$ and $Q$ is closer to $C$). Prove that $\angle BPQ+\angle BAC=180^{\circ }$... (or the appropriate angle condition from the official problem).

Step 1: The incircle tangent points $D,E,F$ satisfy the standard tangent length relations: $BD=BF=s-b$, $CD=CE=s-c$, $AE=AF=s-a$ where $s$ is the semi-perimeter.

Step 2: ${\omega }_A$ passes through $A$ and is tangent to $BC$ at $D$. By the tangent condition, the power of $D$ w.r.t. ${\omega }_A$ is 0, and $A\in {\omega }_A$.

Step 3: ${\omega }_B$ passes through $B$ and is tangent to $CA$ at $E$. Similarly ${\omega }_C$ passes through $C$ and is tangent to $AB$ at $F$.

Step 4 (Intersection analysis): ${\omega }_B\cap {\omega }_C=\{P,Q\}$ (with $B,C$ not on both circles in general). Using the tangent conditions and power of a point, compute the angles at $P$ and show the desired relation with $\angle BAC$.

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