If S is a point on side PQ of a ΔPQR such that PS = QS = RS, then ______.

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KollegeApply · Accepted Answer

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)] $\Rightarrow$ $\angle$1 = $\angle$2 ...(ii) [Angles opposite to equal sides are equal] Similarly, in ∆RSQ, RS = SQ $\Rightarrow$ $\angle$3 = $\angle$4 ...(iii) [Angles opposite to equal sides are equal] Now, in ∆PQR, Sum of angles = $180^\circ$ $\Rightarrow$ $\angle$P + $\angle$Q + $\angle$P = $180^\circ$ $\Rightarrow$ $\angle$2 + $\angle$4 + $\angle$1 + $\angle$3 = $180^\circ$ $\Rightarrow$ $\angle$1 + $\angle$3 + $\angle$1 + $\angle$3 = $180^\circ$ $\Rightarrow$ 2($\angle$1 + $\angle$3) = $180^\circ$ $\Rightarrow$ $\angle$1 + $\angle$3 = 180^circ/2 = $90^\circ$ $\therefore$ $\angle$R = $90^\circ$ In ∆PQR, by Pythagoras theorem, PR2 + QR2 = PQ2 [Solution diagram: 270217_solution_1.png]

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