Prove converse of Pythagoras theorem!
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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Answer:
Converse of Pythagorean theorem is defined as that "If square of a side is equal to sum of square of other two sides then triangle must be right angle triangle."
Before proving the converse of Pythagorean theorem, we have to assume that Pythagorean theorem is already proved.
If length of sides of triangles are a, b and c and c2 = a2 + b2, then triangle must be right angle.
Pythagorean Theorem Test
Construct a another triangle, △EGF, such as AC = EG = b and BC = FG = a.
Converse of Pythagorean Theorem
In △EGF,
By Pythagoras Theorem
EF2 = EG2 + FG2 = b2 + a2 ............(1)
In △EGF,
By Pythagoras Theorem
AB2 = AC2 + BC2 = b2 + a2 ............(2)
From equation (1) and (2), we have
EF2 = AB2
EF = AB
⇒ △ ACB ≅ △EGF (By SSS postulate)
⇒ ∠G is right angle
Thus, △EGF is a right triangle.
Hence, we can say that converse of Pythagorean theorem also hold.
Hence Proved.