Question

show that in a quadrilateral ABCD, a b + BC + CD + D a smaller than <2 (BD + AC)​

Answer 1

Step-by-step explanation:

• Given any quadrilateral ABCD
• Let the diagonals AC and BD intersect eachother at O.
• In ∆OAB, OA+OB > AB (sum of 2 sides of a triangle > the 3rd side) ....(i)
• In ∆OBC, OB+OC > BC .....(ii)
• In ∆OCD, OC+OD > CD .....(iii)
• In ∆ODA, OD+OA > DA ......(iv)
• On adding (i),(ii),(iii),(iv), we get
• 2(OA+OC+OB+OD) > AB+BC+CD+DA
• 2(AC+BD) > AB+BC+CD+DA
• Hence, AB+BC+CD+DA < 2(AC+BD) (proved).