Question

In the given pentagon ABCDE, prove that 2CD < Perimeter (ABCDE).

Answer 1

In triangle ABC, AC < AB + BC ------------------- 1
In triangle ADE, AD < AE + DE ------------------- 2
Adding 1 and 2,
AC + AD < AB + BC + AE + DE ----------------3
In triangle ADC, CD < AD + AC ------------------- 4
From 3 and 4, we can write
CD < AB + BC + AE + DE
Adding CD to both sides,
2CD < AB + BC + CD + DE + EA
Or, 2 CD < Perimeter of ABCDE