In a triangle PQR, ∠P = 90° and PS is perpendicular to QR. Then prove that {1\over PS²}- {1\over PQ²}= {1\over PR²}
Let PQ = b, PR = c, QR = a
Let QS = m and SR = n, so that m + n = a.
Since PS ⟂ QR, the triangles PQS, PRS, and PQR are similar.
From the property of similar triangles:
QS / PQ = PQ / QR
⇒ QS = (PQ)² / QR
⇒ m = b² / a
Similarly,
SR / PR = PR / QR
⇒ SR = (PR)² / QR
⇒ n = c² / a
We know that the square of the perpendicular from the right angle to the hypotenuse is equal to the product of the segments it divides the hypotenuse into.
So, PS² = QS × SR
⇒ PS² = (b² / a) × (c² / a)
⇒ PS² = (b² c²) / a²
Therefore, 1 / PS² = a² / (b² c²)
From Pythagoras theorem: a² = b² + c²
Substitute this value:
1 / PS² = (b² + c²) / (b² c²)
⇒ 1 / PS² = 1 / b² + 1 / c²
⇒ 1 / PS² – 1 / PQ² = 1 / PR² (Hence proved)
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