Question

Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

Answer 1

with respect to the image

triangle AOB = OA + OB > AB ( TRIANGLE INEQUALITY)

Triangle BOC = OB + OC > BC

IN TRIANGLE COD = OC + OD > CD

IN TRIANGLE AOD = OD + DA >AD

(( ADD THE INEQUALITIES))

2(OA+OB+OC+OD)>AB+BC+CD+AD

WITH RESPECT TO FIGURE

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

HENCE PROVED

Answer 2

Answer:

Refer to the attachment above. ⬆️

hope it helps!!