(i) As we have a^2+b^2=c^2, which means that sum of the squares of the two sides a and b is equal to square on the third side c. Hence, /_C opposite side c will be right angle.
![enter image source here]()
Assume, it is not so, then draw a perpendicular from A to BC, let it be at C'. Now according to Pythagoras theorem, a^2+b^2=(AC')^2. Hence, AC'=c=AC. But this is not possible. Hence, /_ACB is a right angle and Delta ABC is a right angled triangle.
Let us recall the cosine formula for triangles, which states that c^2=a^2+b^2-2abcosC.
(ii) As range of /_C is 0^@< C< 180^@, if /_C is obtuse cosC is negative and hence c^2=a^2+b^2+2ab|cosC|. Hence, a^2+b^2< c^2 means /_C is obtuse.
Let us use Pythagoras theorem to check it and draw DeltaABC with /_C>90^@ and draw AO perpendicular on extended BC as shown. Now according to Pythagoras theorem
![enter image source here]()
a^2+b^2=BC^2+AC^2
= (BO-OC)^2+AC^2
= BO^2+OC^2-2BOxxCO+AO^2+OC^2
= BO^2+AO^2-2OC(BO-OC)
= AB^2-2OCxxBC=c^2-OCxxBC
Hence a^2+b^2< c^2
(iii) and if /_C is acute cosC is positive and hence c^2=a^2+b^2-2ab|cosC|. Hence, a^2+b^2> c^2 means /_C is acute.
Again using Pythagoras theorem to check this, draw DeltaABC with /_C<90^@ and draw AO perpendicular on BC as shown. Now according to Pythagoras theorem
![enter image source here]()
a^2+b^2=BC^2+AC^2
= (BO+OC)^2+AO^2+OC^2
= BO^2+OC^2+2BOxxCO+AO^2+OC^2
= AB^2+2OC(CO+OB)
= c^2+2axxOC
Hence a^2+b^2> c^2
