Question

In a quadrilateral ABCD, AC and BD are diagonals. 

Prove that 

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

Answer 1

Given: In quadrilateral ABCD, AC and BD are the diagonals.

To Prove:

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

Proof:

Step 1: Prove (i) AB + BC + CD > AD

Consider triangle ABD.

By the triangle inequality, in triangle ABD: AB + BD > AD   ...(1)

Consider triangle BCD.

By the triangle inequality, in triangle BCD: BC + CD > BD   ...(2)

Adding (1) and (2): AB + BD + BC + CD > AD + BD

Cancel BD on both sides: AB + BC + CD > AD

Hence proved i.

Step 2: Prove (ii) AB + BC + CD + DA > 2AC

In triangle ABC, by triangle inequality: AB + BC > AC   ...(3)

In triangle ADC, by triangle inequality: CD + DA > AC   ...(4)

Adding (3) and (4): AB + BC + CD + DA > AC + AC

⇒ AB + BC + CD + DA > 2AC

Hence proved ii.

Step 3: Prove (iii) AB + BC + CD + DA > 2BD

In triangle ABD, by triangle inequality: AB + AD > BD   ...(5)

In triangle BCD, by triangle inequality: BC + CD > BD   ...(6)

Adding (5) and (6): AB + AD + BC + CD > BD + BD

⇒ AB + BC + CD + DA > 2BD

Hence proved iii.

Step 4: Prove (iv) AB + BC + CD + DA > AC + BD

From step 2: AB + BC + CD + DA > 2AC

From step 3: AB + BC + CD + DA > 2BD

Since AB + BC + CD + DA > 2AC and also > 2BD, it must be greater than the sum AC + BD because AC + BD < 2AC or 2BD since both AC and BD are positive sides.

Alternatively,

Using triangle inequality in triangles ABC and ADC: AB + BC > AC, CD + DA > AC, So, AB + BC + CD + DA > 2AC > AC + BD if AC, BD positive numbers.

And similarly for BD, by inequality properties.

This proves: AB + BC + CD + DA > AC + BD

Hence proved iv.

We have proved that in quadrilateral ABCD with diagonals AC and BD,

AB + BC + CD > AD

AB + BC + CD + DA > 2AC

AB + BC + CD + DA > 2BD

AB + BC + CD + DA > AC + BD