1. **Problem statement:** Given an acute triangle $ABC$ with $AB < AC$, and a variable point $P$ on side $BC$. Define $\omega_B$ as the circle passing through $P$ and tangent to $AB$ at $B$, and $\omega_C$ as the circle passing through $P$ and tangent to $AC$ at $C$. Let $Q$ be the second intersection of $\omega_B$ and $\omega_C$. Prove that the line $PQ$ passes through a fixed point independent of $P$. 2. **Key concepts and formulas:** - The circle tangent to a line at a point and passing through another point is uniquely defined. - Power of a point and radical axis properties. - The radical axis of two circles is the locus of points with equal power to both circles. - The intersection of two circles lies on their radical axis. 3. **Step-by-step solution:** **Step 1:** Consider the two circles $\omega_B$ and $\omega_C$. - $\omega_B$ passes through $P$ and is tangent to $AB$ at $B$. - $\omega_C$ passes through $P$ and is tangent to $AC$ at $C$. **Step 2:** Since $\omega_B$ is tangent to $AB$ at $B$, the radius at $B$ is perpendicular to $AB$. Similarly, $\omega_C$ is tangent to $AC$ at $C$, so its radius at $C$ is perpendicular to $AC$. **Step 3:** Let $O_B$ and $O_C$ be the centers of $\omega_B$ and $\omega_C$, respectively. - $O_B$ lies on the line perpendicular to $AB$ at $B$. - $O_C$ lies on the line perpendicular to $AC$ at $C$. **Step 4:** Since $P$ lies on both circles, the distances $O_BP = O_BB$ and $O_CP = O_CC$ hold. **Step 5:** The radical axis of $\omega_B$ and $\omega_C$ is the line through $Q$ and $P$ because $Q$ is the second intersection point of the two circles. **Step 6:** The radical axis is the locus of points with equal power with respect to both circles. **Step 7:** Consider the fixed point $A$. - Since $\omega_B$ is tangent to $AB$ at $B$, the power of $A$ with respect to $\omega_B$ is zero because $A$ lies on $AB$ and the circle is tangent at $B$. - Similarly, the power of $A$ with respect to $\omega_C$ is zero because $A$ lies on $AC$ and $\omega_C$ is tangent at $C$. **Step 8:** Therefore, $A$ has equal power with respect to both circles, so $A$ lies on the radical axis of $\omega_B$ and $\omega_C$. **Step 9:** Since $P$ and $Q$ lie on the radical axis, and $A$ also lies on it, the line $PQ$ passes through the fixed point $A$ regardless of the position of $P$ on $BC$. **Final conclusion:** The line $PQ$ always passes through the fixed point $A$, independent of $P$. **Answer:** The fixed point through which $PQ$ passes is the vertex $A$ of the triangle.