Q23 of 46 Page 1

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right-angle.


Given: A triangle ABC in which AC2 = AB2 + BC2


To prove: ∠B = 90°



Construction: Draw ΔPQR right angles at Q, such that


PQ = AB and QR = BC


Proof:


In ΔPQR, Using Pythagoras Theorem


(Hypotenuse)2 = (Perpendicular)2 + (Base)2


PR2 = PQ2 + QR2


Since, PQ = AB and QR = BC (By construction)


⇒ PR2 = AB2 + BC2 …(i)


Also, given that


AC2 = AB2 + BC2 …(ii)


From eq. (i) and (ii), we have


PR2 = AC2


⇒ PR = AC …(iii)


In ΔABC And ΔPQR


AC = PR [from (iii)]


AB = PQ [by construction]


BC = RQ [by construction]


∴ ΔABC ≅ ΔPQR [by SSS criterion]


⇒ ∠B = ∠Q


Since, ∠Q = 90°


∴ ∠B = 90°


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