Question

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Answer 1

Given: A right ΔABC right angled at B

To prove : AC² = AB² + BC²

Construction: Draw AD ⊥ AC

Proof: ΔADB and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC  (common)

∴ ΔADB ∼ ΔABC  (by AA similarly criterion)

`=> (AD)/(AB) = (AB)/(AC)`

⇒ AD × AC = AB²    ...... (1)

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)

`=> (CD)/(BC) = (BC)/(AC)`

⇒ 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

Hence Proved.