what is the proof for pythagoras theorem?

Dear Student!

Here is the answer to your query.

Given : A right ΔABC right angled at B

To prove : AC² = AB² + BC²

Construction : Draw AD ⊥ AC

Proof : ΔABD and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC  (common)

∴ ΔADB ∼ ΔABC  (by AA similarly criterion)

⇒ AD × AC = AB²    ...... (1)

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)

⇒ CD × AC = BC²    ........ (2)

Adding (1) and (2) we get

AB² + BC² = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC²

∴ AC² = AB² + BC²

Cheers!