Answer: (i) $$PR^2 = PS^2 + QR \times ST + \left(\frac{QR}{2}\right)^2$$ (ii) $$PQ^2 = PS^2 - QR \times ST + \left(\frac{QR}{2}\right)^2$$ $$QS = SR = \frac{QR}{2}$$ ...(1) (S is the midpoint of side QR) In $$\triangle PTS$$, $$\angle T = 90°$$ According to Pythagoras theorem, $$PS^2 = PT^2 + ST^2$$ ...(2) (i) In $$\triangle PTR$$, $$\angle T = 90°$$ According to Pythagoras theorem, $$PR^2 = PT^2 + RT^2$$ $$\therefore PR^2 = PT^2 + (ST + SR)^2$$ $$\therefore PR^2 = PT^2 + (ST + \frac{QR}{2})^2$$ $$\therefore PR^2 = PT^2 + ST^2 + 2ST \times \left(\frac{QR}{2}\right) + \left(\frac{QR}{2}\right)^2$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$ [(a + b)^2 = a^2 + 2ab + b^2] $$\therefore PR^2 = (PT^2 + ST^2) + ST \times QR + \left(\frac{QR}{2}\right)^2$$ $$\therefore PR^2 = PS^2 + ST \times QR + \left(\frac{QR}{2}\right)^2$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$ [From 2] (ii) In $$\triangle PTQ$$, $$\angle T = 90°$$ According to Pythagoras theorem, $$PQ^2 = PT^2 + TQ^2$$ $$\therefore PQ^2 = PT^2 + (QS - ST)^2$$ $$\therefore PQ^2 = PT^2 + \left(\frac{QR}{2} - ST\right)^2$$ $$\therefore PQ^2 = PT^2 + \left(\frac{QR}{2}\right)^2 - 2 \times \frac{QR}{2} \times ST + ST^2$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$ [(a - b)^2 = a^2 - 2ab + b^2] $$\therefore PQ^2 = (PT^2 + ST^2) - QR \times ST + \left(\frac{QR}{2}\right)^2$$ $$\therefore PQ^2 = PS^2 - QR \times ST + \left(\frac{QR}{2}\right)^2$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$ [From 2]