Question

In a quadrilateral ABCD. Prove that:

1. BC + CD + DA > AB 
2. AB + BC + CD + DA > 2AC 
3. AB + BC + CD + DA > 2BD

Answer 1

Given:

• ABCD is any quadrilateral, AC and BD are its diagonals.

To Prove:

1. BC + CD + DA > AB 
2. AB + BC + CD + DA > 2AC 
3. AB + BC + CD + DA > 2BD

Proof [Step-wise]:

a. BC + CD + DA > AB

1. In triangle ACD, by the triangle inequality, CD + DA > AC.
2. Therefore, BC + CD + DA > BC + AC.   ...(Add BC to both sides)
3. In triangle ABC, by the triangle inequality, BC + AC > AB.
4. Combining (2) and (3) gives BC + CD + DA > AB.   ...(Transitive)

b. AB + BC + CD + DA > 2AC

1. In triangle ABC, AB + BC > AC.
2. In triangle ADC, AD + DC > AC.
3. Add the two inequalities: (AB + BC) + (AD + DC) > AC + AC = 2AC
4. Hence, AB + BC + CD + DA > 2AC.

c. AB + BC + CD + DA > 2BD

1. In triangle ABD, AB + AD > BD.
2. In triangle BCD, BC + CD > BD.
3. Add the two inequalities: (AB + AD) + (BC + CD) > BD + BD = 2BD.
4. Hence, AB + BC + CD + DA > 2BD.

All three inequalities hold for any quadrilateral ABCD:

1. BC + CD + DA > AB,
2. AB + BC + CD + DA > 2AC, 
3. AB + BC + CD + DA > 2BD.