Pythagoras theoremIn a right triangle the square of the hypotenuse is equal to the square of other two sides Given -ABC is a right triangle Right angled at BTo prove -AC2=AB2+BC2Construction -BD prepndicular ACProof- In triangle ABC and triangle ABD Angle A is common Angle ADB=angle B (both common) Therefore triangle ABM similar to triangle ABC AM/BC=BC/AC BC2=AM×AC ……eq1 Similarly, in triangle BMC & triangle ABC CM/AB =AB/AC AB2=CM×AC……..eq 2 From 1 and 2AB2+BC2=AM×AC+CM×ACAB2+BC2=AC(AM +MC)AB2+BC2=AC×ACAB2+BC2=AC2(Proved)