Question

state and prove converse of pythogras theorem​

Answer 1

Answer:

In a triangle , if square of one side is equal to the sum of the squares of the other two sides , then the angle opposite the first side is a right angle.

TO PROVE : ∠C =90

PROOF:

GIven a ΔABC, with BC = a, AC = b and AB = c.

Also given, c2 = a2 + b2  --- (1)

Now construct a right angled ΔDEF, with sides EF = BC = a, AC = DF = b.

Let DE = d and ∠EFD = 900.

SInce, ΔDEF is a right angled triangle, we can use Pythagoras theorem,  

⇒ d2 = a2 + b2

But by (1),  c2 = a2 + b2

Therefore, c = d

i.e. AB = DE

Thus, by construction , By SSS test, ΔABC ≃ ΔDEF

Thus, ΔABC is a right angled triangle with ∠ACB = 90.

Answer 2

Step-by-step explanation:

Statement:

In a Triangle the square of longer side is equal to the sum of squares of the other two sides, then the triangle is a right angled triangle.

Given -

A Triangle ABC such that

BC² = AB² + AC²

To Prove -

Angle A = 90°

Construction -

Draw a ∆DEF such that AB = DE and AC = DF and Angle D = 90°

Proof -

In ∆ABC,

BC² = AB² + AC² - Given

In ∆ DEF

EF² = DE² + DF²

Therefore,

EF² = AB² + AC²

(Since AB = DE, AC = DF)

Therefore,

BC² = EF² ie - BC = EF

Now, In ∆ABC and ∆DEF

AB = DE - By Construction

AC = DF - By Construction

BC = EF

Therefore

∆ABC ≅ ∆DEF by SSS test.

Thus,

Angle A = Angle D - CPCT

But, Angle D = 90° ( As per construction)

Therefore

Angle A = 90°

Hence Proved!