Question
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

Answer

Given ABCD is a quadilateral,

Construction join diagonais AC and BD.

To show AB + BC + CD + DA > AC + BD 

In $\triangle\text{ABC},$ AB + BC > AC [Sum of two sides of a triangle is greater than the third side] ...(i)

In $\triangle\text{BCD},$ BC + CD > BD [Sum of two sides of a triangle is greater than the third side] ...(ii)

In $\triangle\text{CDA},$ CD + DA > AC [Sum of two sides of a triangle is greater than the third side] ...(iii)

In $\triangle\text{DAB},$ DA + AB + > BD [Sum of two sides of a triangle is greater than the third side] ...(iv)

On adding Eqs. (i), (ii), (iii) (iv), We get

2(AB + BC + CD + DA) > 2 (AC + BD)

⇒ AB + BC + CD + DA > AC + BD

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