Question

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

Figure

Answer 1

Sum of any two sides of a triangle is greater than the third side.

In $△$ABC:
AB + BC > AC

In $△$ADC:
CD + DA > AC

Adding the above two:

AB + BC + CD + DA > 2 AC ... (i)

In $△$ADB:
AD + AB > BD

In $△$BDC:
CD + BC > BD

Adding the above two:

AB + BC + CD + DA >2 BD ... (ii)

Adding equation (i) and (ii):

AB + BC + CD + DA+AB + BC + CD + DA> 2(AC+BD)
=> 2(AB + BC + CD + DA)>2(AC+BD)
=> AB + BC + CD + DA > AC+BD