Question

how can we prove pythagoras theorem

Answer 1

Pythagoras theorem is :
     In a right angle triangle , sum of squares of sides is equal to the square of the hypotenuse.

We can solve that using similar triangles principles.  This is the technique that Pythagoras used.

See the diagram enclosed.

ABC is a right angle triangle at B and BD is a perpendicular from B on to AC, the diagonal.

Triangles ABC and BDC are similar, as:
     . angle DBC = 90 - angle C = angle A
     . angle ABC = angle BDC = 90

Ratios of corresponding sides are equal.
     AB / BD = BC / DC = AC / BC
    So we get  BC² = AC * DC     --- (1)

Triangle ABC amd ADB are similar, as:
    . angle DBA = 90 - (90 - C) = angle C
    . angle ABC = 90 = angle ADB

so ratios of corresponding sides are equal
         AB / AD = BC / BD = AC / AB 
       So we get  AB² = AD * AC  --- (2)

  Add (1) and (2) to get:
       AB² + BC² = AC * ( AD + DC) = AC * AC = AC²

So the theorem is proved.

Answer 2

Step-by-step explanation:

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. ...................................................