Question

If S is a point on side PQ of a triangle PQR such that PS = QS = RS, then prove PR^2+QR^2=PQ^2

Answer 1

Let PS = QS = RS = x. Then PQ = 2x

Let ∠PSR = θ. Then ∠QSR = 180 - θ

Using Cosine rule in △PRS
PR² = PS² + RS² - 2 (PS)(RS) cos(∠PSR)
PR² = x² + x² - 2 (x)(x) cos(θ)
PR² = 2x² - 2x²cos(θ)

Using Cosine rule in △QRS
QR² = QS² + RS² - 2 (QS)(RS) cos(∠QSR)
QR² = x² + x² - 2 (x)(x) cos(180-θ)
QR² = 2x² - 2x²cos(180-θ)

PR² + QR² = (2x² - 2x²cos(θ)) + (2x² - 2x²cos(180-θ))
PR² + QR² = 4x² - 2x² (cos(θ) + cos(180-θ))
PR² + QR² = 4x² - 2x² (cos(θ) + (-cos(θ)))
PR² + QR² = 4x²
PR² + QR² = (2x)²
PR² + QR² = PQ²