In the figure 2.35, \(\triangle PQR\) is an equilateral triangle. Point S is on seg QR such that \(QS = \frac{1}{3} QR\). Prove that: \(9\,PS^2 = 7\,PQ^2\)
Figure: PT ⊥ QR with Q–S–T–R.
Answer: With side a, PT² = \(\frac{3a^2}{4}\) and ST = \(\frac{a}{6}\), so PS² = \(\frac{7a^2}{9}\), i.e. \(9PS^2 = 7PQ^2\).
Let PQ = QR = PR = a. In an equilateral triangle the perpendicular PT bisects QR, so QT = \(\frac{a}{2}\), and
\[PT^2 = PQ^2 - QT^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}\]QS = \(\frac{a}{3}\), so ST = QT − QS = \(\frac{a}{2} - \frac{a}{3} = \frac{a}{6}\).
In right-angled \(\triangle PTS\):
\[PS^2 = PT^2 + ST^2 = \frac{3a^2}{4} + \frac{a^2}{36}\] \[= \frac{27a^2 + a^2}{36} = \frac{7a^2}{9}\] \[\therefore 9PS^2 = 7a^2 = 7PQ^2\]Hence proved.
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