Question
ABCD is quadrilateral.
Prove that (AB + BC + CD + DA) > (AC + BD)

Answer


Given, ABCD is a quadrilateral AC and BD are joined.
Proof : Now in $\triangle\text{ABC},$
AB + BC > AC...(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly in $\triangle\text{ADC},$
AC + CD > AC...(ii)
In $\triangle\text{ABD},$
AB + AD > BD...(iii)
and in $\triangle\text{BCD},$
BC + CD > BD...(iv)
Adding (i), (ii), (iii) and (iv)
AB + BC + CD + AD + AB + AD + BC + CD > AC + AC + BD + BD
⇒ 2(AB + BC + CD + AD) > 2(AP + BD)
⇒ AB + BC + CD + AD > AC + BD
Hence proved.

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