Question

show that in a quadrilateral ABCD AB+BC+CD+DA<2(BD+AC)

Answer 1

ABCD is a quadrilateral and AC, and BD are the diagonals.
Sum of the two sides of a triangle is greater than the third side.
So, considering the triangle ABC, BCD, CAD and BAD, we get
AB + BC > AC
CD + AD > AC
AB + AD > BD
BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)
⇒ 2(AB + BC + CA + AD) > 2(AC + BD)
⇒ (AB + BC + CA + AD) > (AC + BD)
⇒ (AC + BD) < (AB + BC + CA + AD)

Answer 2

hope it would help u .......