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