Question

In the quadrilateral ABCD, prove that AB + BC + CD + AD > AC + BD.

Answer 1

Given: Quadrilateral ABCD with diagonals AC and BD.

To prove: AB + BC + CD + AD > AC + BD

Proof (triangle-inequality + averaging)

Apply the triangle inequality in four triangles that share the diagonals:

• In ΔABC: AB + BC > AC
• In ΔCDA: CD + DA > AC

Add these two:

AB + BC + CD + AD > 2AC   ...(1)

• In ΔBCD: BC + CD > BD
• In ΔDAB: DA + AB > BD

Add these two:

AB + BC + CD + AD > 2BD   ...(2)

Now add (1) and (2) and divide by 2:

AB + BC + CD + DA > `(2AC + 2BD)/2` = AC + BD

Hence, AB + BC + CD + AD > AC + BD, as required.