Converse of Pythagoras theorem? Proof?
Answers
In a triangle, if the square of one side is equal to the sum of square of other two sides then prove that the triangle is right angled triangle.
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Given : AC² = AB² + BC²
To prove : ABC is a right angled triangle.
Construction : Draw a right angled triangle PQR such that, angle Q = 90°, AB = PQ, BC = QR.
Proof : In triangle PQR,
Angle Q = 90° ( by construction )
Also,
PR² = PQ² + QR² ( By using Pythagoras theorem )...(1)
But,
AC² = AB² + BC² ( Given )
Also, AB = PQ and BC = QR ( by construction )
Therefore,
AC² = PQ²+ QR²....(2)
From eq (1) and (2),
PR² = AC²
So, PR = AC
Now,
In ∆ABC and ∆PQR,
AB = PQ ( By construction )
BC = QR ( By construction )
AC = PR ( Proved above )
Hence,
∆ABC is congruent to ∆PQR by SSS criteria.
Therefore, Angle B = Angle Q ( By CPCT )
But,
Angle Q = 90° ( By construction )
Therefore,
Angle B = 90°
Thus, ABC is a right angled triangle with Angle B = 90°
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Hence proved!
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In triangle, if (longer side)² = Sum of the squares of other side then the triangle is right angled triangle.
Proof in attachment
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...Hope it helps....
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