Question

In a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides ?
prove we are given a right triangle ABC right angled at B.

Answer 1

Given: A right triangle ABC right angled at B.

To Prove: AC2 = AB2 + BC2

Construction: Draw BD ⊥ AC

Proof:

In Δ ADB and Δ ABC,

∠ ADB = ∠ ABC (each 90°)

∠ BAD = ∠ CAB (common)



Δ ADB ~ Δ ABC (By AA similarity criterion)

Now, AD/AB = AB/AC (corresponding sides are proportional)

AB2 = AD × AC … (i)

Similarly, Δ BDC ~ Δ ABC

BC2 = CD × AC … (ii)

Adding (i) and (ii)

AB2 + BC2 = (AD × AC) + (CD × AC)

AB2 + BC2 = AC × (AD + CD)

AB2 + BC2 = AC2

Hence Proved.

Answer 2

Given : A right triangle ABC right angled at B.

To prove : AC^2 = AB^2 + BC^2

Construction : Draw BD ⊥ AC

Proof : In ΔADB and Δ ABC

∠ADB = ∠ABC (each 90°)

∠BAD =∠CAB (common)

ΔADB ~ ΔABC (By AA similarity criterion)

AD  ÷ AB  =  AB  ÷ AC (corresponding sides are proportional)

⇒ AB^2 = AD × AC …(i)

Similarly ΔBDC ~ ΔABC

⇒ BC^2 = CD × AC …(ii)

Adding (1) and (2)

AB^2 + BC^2 = AD × AC + CD × AC

⇒ AB^2 + BC^2 = AC × (AD + CD)

⇒ AB^2 + BC^2 = AC^2