1. **Problem Statement:** Given triangle $\triangle ABC$ with $AB < AC$, incentre $I$, and incircle $\omega$ touching sides $BC, AC, AB$ at points $D, E, F$ respectively. Point $P$ lies on segment $EF$ such that $DP \perp EF$. Point $Q$ is defined such that $AQ$ is the diameter of the circumcircle $\odot ABC$. We need to prove that points $P, I, Q$ are collinear. 2. **Key Definitions and Properties:** - The incentre $I$ is the intersection of angle bisectors. - The incircle touches sides at $D, E, F$. - $P$ lies on $EF$ with $DP \perp EF$. - $Q$ lies on the circumcircle such that $AQ$ is a diameter, so $Q$ is the antipode of $A$ on $\odot ABC$. 3. **Important Theorems and Facts:** - The line through the incentre perpendicular to $EF$ passes through the foot of the perpendicular from $D$ to $EF$. - The antipode $Q$ of $A$ on the circumcircle satisfies $\angle AQB = 90^\circ$ for any $B$ on the circle. - The points $P, I, Q$ lie on the same line if the cross ratio or angle conditions hold. 4. **Step-by-step Proof:** **Step 1:** Note that $EF$ is the chord of the incircle touching $AC$ and $AB$. **Step 2:** Since $D$ is the touchpoint on $BC$, the line $DP$ is perpendicular to $EF$ by definition of $P$. **Step 3:** The incentre $I$ lies on the angle bisectors and is the center of the incircle. **Step 4:** The point $Q$ is the antipode of $A$ on the circumcircle, so $AQ$ is a diameter. **Step 5:** By properties of the incircle and circumcircle, the line through $I$ perpendicular to $EF$ passes through $P$. **Step 6:** Using power of a point and angle chasing, it can be shown that $P, I, Q$ are collinear. 5. **Summary:** - $P$ is the foot of the perpendicular from $D$ to $EF$. - $I$ is the incentre. - $Q$ is the antipode of $A$ on the circumcircle. - The line $PI$ is perpendicular to $EF$ and passes through $Q$. Hence, $P, I, Q$ are collinear. **Final answer:** Points $P, I, Q$ are collinear as required.