I need help with this proof

Question

Given: Quad. ABCD

Prove: AB + BC + CD + DA + > 2(AC)

Answer 1

The diagonal AC splits the quadrilateral into two triangles: ΔABC and ΔACD.  The Triangle Inequality Theorem says that "Any side of a triangle must be shorter than the other two sides added together. So for ΔABC:

 

AB + BC > AC

 

For triangle ΔACD:

 

AD + CD > AC

 

Add the two inequalities together:

 

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC