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6
Theorem-6.9: Converse of Pythagoras theoremm
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Statement: In a triangle, if square of one side is equal to the sum of the
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squares of the other two sides, then the angle opposite the first side is a right
s q u a r e s o f f t t h e o t t h e r t t w o s i i d e s , , t t h e n t t h e a n g l l e o p p o s i i t t e t t h e f f i i r s t t s i i d e i i s a r i i g h t t
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angle.
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Given: A ABC in which ACC = AB + BC 2
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To prove: ABC = 90 o
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Construction: Construct a PQR in which Q = 90 , PQ = AB and QR =
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Proof: In ABC,
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AC = AB + BC (i) (given)
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In PQR, Q = 90 (by construction))
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Therefore, using Pythagoras theorem, we get,,
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PR = PQ + QR 2

As PQ = AB and QR = BC (by construction)

Therefore, we get,
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2

PR = AB + BC (ii)
From (i) and (ii), we get,
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AC = PR
AC = PR
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Now in ABC and PQR, wee have,
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AB = PQ, BC = QR (by construction))
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and AC = PR (proved above))
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Therefore, by SSS criterion of congruency, we get,,
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ABC PQR

Therefore, B = Q (By cpct)

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Since, Q = 90
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B = 90
Hence, proved.

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