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Question
If S is a point on side PQ of a ΔPQR such that PS = QS = RS, then ______.
Options
PR . QR = RS2
QS2 + RS2 = QR2
PR2 + QR2 = PQ2
PS2 + RS2 = PR2
Solution
If S is a point on side PQ of a ΔPQR such that PS = QS = RS, then PR2 + QR2 = PQ2.
Explanation:
Given, in ∆PQR,
PS = QS = RS ...(i)
In ∆PSR,
PS = RS ...[From equation (i)]
⇒ ∠1 = ∠2 ...(ii) [Angles opposite to equal sides are equal]
Similarly, in ∆RSQ,
RS = SQ
⇒ ∠3 = ∠4 ...(iii) [Angles opposite to equal sides are equal]
Now, in ∆PQR,
Sum of angles = 180°
⇒ ∠P + ∠Q + ∠P = 180°
⇒ ∠2 + ∠4 + ∠1 + ∠3 = 180°
⇒ ∠1 + ∠3 + ∠1 + ∠3 = 180°
⇒ 2(∠1 + ∠3) = 180°
⇒ ∠1 + ∠3 = `180^circ/2` = 90°
∴ ∠R = 90°
In ∆PQR, by Pythagoras theorem,
PR2 + QR2 = PQ2
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