Question

Heyyyyyyy!! Proof of Converse of Pythagoras theorem???!?

Answer 1

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

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°

____________________

Hence proved!

Answer 2

Answer:

In a right angled trianglPythagoras’ Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

As shown in a right angled triangle ABC above,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

(AC)2 = (BC)2 + (AB)2

That is, in ΔABC, if c2=a2+b2 then ∠C is a right triangle, ΔPQR being the right angle.

We can prove this by contradiction.

Let us assume that c2=a2+b2 in ΔABC and the triangle is not a right triangle.

Now consider another triangle ΔPQR. We construct ΔPQR so that PR=a, QR=b and ∠R is a right angle.

By the Pythagorean Theorem, (PQ)2=a2+b2.

But we know that a2+b2=c2 and a2+b2=c2 and c=AB.

So, (PQ)2=a2+b2=(AB)2.

That is, (PQ)2=(AB)2.

Since PQ and AB are lengths of sides, we can take positive square roots.

PQ=AB

That is, all the three sides of ΔPQR are congruent to the three sides of ΔABC. So, the two triangles are congruent by the Side-Side-Side Congruence Property.

Since ΔABC is congruent to ΔPQR and ΔPQR is a right triangle, ΔABC must also be a right triangle.

Hope this helps ☺️