Q.4 ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?

In \(\triangle \)ABC, we have
AB + BC > AC …(i) [Sum of any two sides is greater than the third side]
In \(\triangle \)BDC, we have
BC + CD > BD …(ii)
In \(\triangle \)ADC, we have CD + DA > AC .... (iii)
In \(\triangle \)DAB, we have DA + AB > BD …(iv)
Adding eq. (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2CD + 2DA > 2AC + 2BD
Or, AB + BC + CD + DA > AC + BD [Dividing both sides by 2]
Hence, proved