Question

Converse of Pythagoras theorem?????? How to prove it?

Answer 1

$\huge\underline\mathscr{Statement}$

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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$\huge\underline\mathscr{Solution}$

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!

Answer 2

Answer:

its simple,,

1. first draw a triangle similar to other, means both the side should be congruent.

2. then prove that hypotenuse of both the triangles are congruent.

3. Then by sss test of congruence both the triangle will be congruent.

4. Then The Angle will be 90° by Corresponding sides of congruent triangle