In the figure 2.28 seg PS is the median of \(\triangle PQR\) and PT ⊥ QR. Prove that,
(1) \(PR^2 = PS^2 + QR \times ST + \left(\dfrac{QR}{2}\right)^2\)
(2) \(PQ^2 = PS^2 - QR \times ST + \left(\dfrac{QR}{2}\right)^2\)
Figure: Points on side QR are in the order Q–T–S–R.
Answer: Apply Pythagoras theorem in \(\triangle PTR\) and \(\triangle PTQ\), writing TR = TS + SR and QT = QS − TS, with QS = SR = QR/2.
S is the midpoint of QR, so \(QS = SR = \frac{QR}{2}\). In right-angled \(\triangle PTS\), \(PT^2 + ST^2 = PS^2\).
(1) In right-angled \(\triangle PTR\), TR = ST + SR:
\[PR^2 = PT^2 + (ST + SR)^2\] \[= PT^2 + ST^2 + 2\,ST \cdot SR + SR^2\] \[= PS^2 + 2 \cdot \frac{QR}{2} \cdot ST + \left(\frac{QR}{2}\right)^2\] \[= PS^2 + QR \times ST + \left(\frac{QR}{2}\right)^2\](2) In right-angled \(\triangle PTQ\), QT = QS − ST:
\[PQ^2 = PT^2 + (QS - ST)^2\] \[= PT^2 + ST^2 - 2\,QS \cdot ST + QS^2\] \[= PS^2 - QR \times ST + \left(\frac{QR}{2}\right)^2\]Hence both results are proved.
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