Question

Prove that in quadrilateral $ABCD,AC<\frac{1}{2}(AB+BC+CD+DA)$

Answer 1

Let $ABCD$ be a quadrilateral.

In $\Delta ABC,AB+BC>AC$ ...(1)(Sum of two sides of a triangle is greater than the third side)

In $\Delta ACD,CD+DA>AC$ ...(2)(Sum of two sides of a triangle is greater than the third side)

Adding $\left(1\right),\left(2\right)$ we get

$AB+BC+CD+DA>AC+AC$

$\Rightarrow AB+BC+CD+DA>2AC$

$\Rightarrow AC<\frac{1}{2}(AB+BC+CD+DA)$