Question

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

Answer 1

Given: A right triangle ABC in which ∠B = 90°

To Prove: (Hypotenuse)² = (Base)² + (perpendicular)²

i.e AC² = AB² + BC²

Construction: From B, Draw BD ⊥ AC

In ΔABC and ΔADB

∠BAC = ∠DAB   [Common]

∠ABC = ∠ADB  [each 90°]

∴ ΔABC ∼ ΔADB  [By AA similarity]

`=> "AB"/"AC" = "AD"/"AB"`

`=> AB² - AD x AC ...(i)

Similarity, ΔABC ∼ ΔBDC

`=> "BC"/"DC" = "AC"/"BC"`

`=> "BC"^2 - "AC" xx "DC"` ...(ii)

On adding (i) and (iii) we get

AB² + BC² = AD x AC + AC x DC

`=>` AB² + BC² = AC(AD + DC)

`=>` AB² + BC² = AC x AC

`=>` AC² = AB² + BC²