In the figure 2.35, ∆ P20QR is an equilatral triangle. Point S is on seg QR such that

Prove that : 9 PS² = 7 PQ²

As, PQR is an equilateral triangle,

Point S is on base QR, such that

PT is perpendicular on side QR from P.

As, PQR is an equilateral triangle we have

PQ = QR = PR [1]

Also, we know that Perpendicular from a vertex to corresponding side in an equilateral triangle bisects the side

Now, In ΔPTQ, By Pythagoras theorem

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ PQ² = PT² + QT²

[2]

As,

We have,

QT - QS = ST

Now, In right angled triangle PST, By Pythagoras theorem

(PS)² = (ST)² + (PT)²

[From 2]

⇒ 36 PS² = 28 PQ²

⇒ 9 PS² = 7 PQ²

Hence Proved.