Question

Proof of converse of Pythagoras theorem with statement.

Answer 1

$\huge\underline\mathfrak\red{Statement}$

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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$\huge\underline\mathfrak\red{Solution}$

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:

plz refer to attachment