*Problem: Prove that the hypotenuse of a right angled triangle is the greatest side.*

**Solution:** **General enunciation**: We have to prove that the hypotenuse of a right angled triangle is the greatest side.

**Particular enunciation**: Let ∆ABC be right angled triangle in which ∠ABC = right angle or 90^{0} and AC be the hypotenuse. It is required to prove that AC be the greatest side.

*Proof*: *We know, in a right angled triangle one angle is right angle and the other two angles are acute angles.*

∴ ∠ABC> ∠ACB

We know, the side opposite to the greater angle is greater.

∴ AC>AB

Again, ∠ABC> ∠BAC

⟹ AC>BC

∴ Side AC be greater than the sides AB and BC.

Therefore, AC be the greatest side. (Proved)

### Problem: Prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.

**Solution:** **General enunciation**: We have to prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.

**Particular enunciation**: Let the greatest side of ∆ABC is AC. It is required to prove that ∠ABC is the greatest angle of the triangle.

Proof: In triangle ABC, AC>AB

We know, the angle opposite to the greater side is greater.

∴ ∠ABC> ∠ACB

Again, AC>BC

⟹∠ABC> ∠BAC

Hence ∠ABC be greater than ∠ACB and ∠BAC.

Therefore, ∠ABC be the greatest angle of the triangle.

*Problem: Prove that difference of any two sides of a triangle is less than the third side.*

**Solution**: General enunciation: We have to prove that difference of any two sides of a triangle is less than the third side.

### Particular enunciation: Let, in ∆ABC, AB is greatest side. Prove that difference of any two sides of a triangle is less than the third side.

**Construction**: By joining C , D.

**Proof**: We know, sum of two side of any triangle is greater than third side.

AC+BC>AB

⟹ AC+BC-AC>AB-AC

⟹BC>AB-AD

⟹BC>BD

⟹BC>BC

So, Difference of any two sides of a triangle is less than the third side.

**(Proved)**