Question

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

Answer 1

Given: AB ⊥ BC

Construction: Draw BE ⊥ AC

To Prove: AB² + BC² = AC²

Proof: In ΔAEB and ΔABC

∠A = ∠A   ...(Common)

∠E = ∠B   ...(Each 90°)

ΔAEB ∼ ΔABC   ...(By AA similarity)

⇒ `"AE"/"AB" = "AB"/"AC"`

⇒ AB² = AE × AC   ...(i)

Now, in ΔCEB and ΔCBA,

∠C = ∠C   ...(Common)

∠E = ∠B   ...(Each 90°)

ΔCEB ∼ ΔCBA   ...(By AA similarity)

⇒ `"CE"/"BC" = "BC"/"AC"`

⇒ BC² = CE × AC   ...(ii)

On adding equations (i) and (ii),

AB² + BC² = AE × AC + CE × AC

⇒ AB² + BC² = AC(AE + CE)

⇒ AB² + BC² = AC × AC

∴ AB² + BC² = AC²

Hence proved.