If S is a point on side PQ of a ∆PQR such that PS = QS = RS, then


Given,

In ∆PQR


PS = QS + RS ……(i)


In ∆PSR


PS = RS ….. [from Equation (i)]


⇒ ∠1 = 2 Equation ….(ii)


Similarly,


In ∆RSQ,


⇒ ∠3 = 4 Equation……(iii)


[Corresponding angles of equal sides are equal]



[By using Equations (ii) and (iii)]


Now in,


∆PQR, sum of angles = 180°


⇒ ∠P + Q + R = 180°


⇒ ∠2 + 4 + 1 + 3 = 180°


⇒ ∠1 + 3 + 1 + 3 = 180°


⇒∠2 (1 + 3) = 180°


⇒ ∠1 + 3 = (180°)/2 = 90°


∴ ∠R = 90°


In ∆PQR, by Pythagoras theorem,


PR2 + QR2 = PQ2

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