OH is the Euler line of ΔPQR and, as such, contains T midway between O and H. Thus, PT ⊥ OH is equivalent to ΔPHO being isosceles, in particular to PO = PH. However, PO = R, the circumradius of ΔPQR. Also, PH = 2R·|cos(∠QPR)| = R implying that PO = PH, as needed.
Concerning PI, we know that the lines PH and PO are isogonal images of each other because PI bisects one of the angles they form. This makes PI either parallel or perpendicular to another bisector of one of the angles formed by PH and PO. When ∠QPR = 120°, PI and PT are distinct and PI ⊥ PT. When ∠QPR = 60°, PI and PT coincide.
