Prove that in a right triangle, the square of the hypotenuse is equal

1.	Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the others two sides.
Answer» Proof: We are given a right triangle ABC right angled at B. We need to prove that AC² = AB² + BC² Let us draw BD ⊥ AC Now, ΔADB ~ ΔABC AD/AB = AB/AC (sides are proportional) AD.AC = AB² … (1) Also, ΔBDC ~ ΔABC CD/BC = BC/AC CD.AC = BC² … (2) Adding (1) and (2) AD.AC + CD.AC = AB² + BC² AC(AD + CD) = AB² + BC² AC.AC = AB² + BC² AC² = AB² + BC²

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

Answer»

Proof:

We are given a right triangle ABC right angled at B.

We need to prove that AC² = AB² + BC²

Let us draw BD ⊥ AC

Now, ΔADB ~ ΔABC

AD/AB = AB/AC (sides are proportional)

AD.AC = AB² … (1)

Also, ΔBDC ~ ΔABC

CD/BC = BC/AC

CD.AC = BC² … (2)

Adding (1) and (2)

AD.AC + CD.AC = AB² + BC²

AC(AD + CD) = AB² + BC²

AC.AC = AB² + BC²

AC² = AB² + BC²