1. **State the problem:** We need to find the size of angle $p$ in a triangle positioned between two parallel lines, given angles of $109^\circ$ and $125^\circ$ at specific points. 2. **Identify key facts:** The two lines are parallel, and the triangle has two equal sides, so it is isosceles. The angle $109^\circ$ is at the top right vertex of the triangle, and the $125^\circ$ angle is an exterior angle adjacent to the base near angle $p$. 3. **Use the property of parallel lines:** The $125^\circ$ angle outside the triangle and the angle adjacent to it inside the triangle are supplementary because they form a linear pair. So, the interior angle adjacent to $125^\circ$ is: $$180^\circ - 125^\circ = 55^\circ$$ 4. **Label the triangle angles:** Let the base angles of the isosceles triangle be $p$ and $55^\circ$ (the one adjacent to $125^\circ$). Since the triangle is isosceles with two equal sides, the base angles are equal, so: $$p = 55^\circ$$ 5. **Check with the triangle angle sum:** The sum of angles in a triangle is $180^\circ$. The angles are $109^\circ$, $p$, and $55^\circ$. Calculate the sum: $$109^\circ + 55^\circ + p = 180^\circ$$ Since $p = 55^\circ$, this holds true: $$109^\circ + 55^\circ + 55^\circ = 219^\circ$$ This is more than $180^\circ$, so our assumption that $p=55^\circ$ is incorrect. Instead, since the triangle is isosceles with two equal sides, the equal angles are the ones opposite those sides. 6. **Reconsider the equal angles:** The two equal sides are the ones adjacent to angle $p$ and the $109^\circ$ angle. Therefore, the angles opposite these sides are equal. So, the angles opposite the equal sides are $p$ and the angle adjacent to $125^\circ$ (which is $55^\circ$). Since these are equal: $$p = 55^\circ$$ 7. **Calculate the third angle:** The third angle is $109^\circ$. 8. **Verify the triangle angle sum:** $$p + p + 109^\circ = 180^\circ$$ $$2p + 109^\circ = 180^\circ$$ $$2p = 180^\circ - 109^\circ = 71^\circ$$ $$p = \frac{71^\circ}{2} = 35.5^\circ$$ **Final answer:** $$p = 35.5^\circ$$