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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 the first side is a right angle.

 

 
 
 
 
 

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In \DeltaABC, the square of the one side is equal to the sum of the square of the other two sides.    AB2 + BC2 = AC2.

To Prove:    \mathrm{\angle B = 90\degree}

Construction: we construct a \DeltaPQR, right-angled at Q such that PQ = AB and QR = BC

Proof:    Now from \DeltaPQR

               PR2 = PQ2 + QR2    [Pythagorus Theorem as \mathrm{\angle Q = 90\degree}]

          or, PR2 = AB2 + BC2    [By construction] -(i)

        But, AC2 = AB2 + BC2    [Given]    -(ii)

          So, AC2 = PR2    [From (i) and (ii)]

           \therefore AC = PR    -(iii)

Now in \DeltaABC and \DeltaPQR

               AB = PQ    [By construction]

               BC = QR    [By construction]

               AC = PR    [Proved in (iii)]

So,    \DeltaABC \cong \DeltaPQR

        Therefore

                        \mathrm{\angle B = \angle Q}    [CPCT]

           But       \mathrm{\angle Q = 90\degree}    [By Construction]

             so,      \mathrm{\angle B = 90\degree},    hence, proved.

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