Question

list 2 ways to prove pythagoras theorem,thank you

Answer 1

first way of proving pythagoras theorem is my practical geometry .another may is my using pigeon hole theorem 

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!

Pythagoras' theorem :-

→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Step-by-step explanation:

It's prove :-

➡ Given :-

→ A △ABC in which ∠ABC = 90° .

➡To prove :-

→ AC² = AB² + BC² .

➡ Construction :-

→ Draw BD ⊥ AC .

➡ Proof :-

In △ADB and △ABC , we have

∠A = ∠A ( common ) .

∠ADB = ∠ABC [ each equal to 90° ] .

∴ △ADB ∼ △ABC [ By AA-similarity ] .

⇒ AD/AB = AB/AC .

⇒ AB² = AD × AC ............(1) .

In △BDC and △ABC , we have

∠C = ∠C ( common ) .

∠BDC = ∠ABC [ each equal to 90° ] .

∴ △BDC ∼ △ABC [ By AA-similarity ] .

⇒ DC/BC = BC/AC .

⇒ BC² = DC × AC. ............(2) .

Add in equation (1) and (2) , we get

⇒ AB² + BC² = AD × AC + DC × AC .

⇒ AB² + BC² = AC( AD + DC ) .

⇒ AB² + BC² = AC × AC .

$\huge \green{ \boxed{ \sf \therefore AC^2 = AB^2 + BC^2 }}$

Hence, it is proved.