Question

Prove that sum of diagonals of a convex pentagon is greater than the perimeter but less than double the perimeter

Answer 1

I'd use the triangles ABC, BCD, CDE, DEA and EAB.

Answer 2

AC>AB, BD>BC, ...
Therefore AC+BD+... > AB+BC+...

Answer 3

But AC<AB+BC, BD<BC+CD, ...
Therefore AC+BD+... < AB+BC+BC+CD+CD+DE+DE+EA+EA+AB = 2*perimeter

Answer 4

thank u

Answer 5

No problem