Question

Question 11
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC).

Answer 1

Construction: Join diagonals AC and BD.
In $\Delta OAB,OA+OB>AB...\left(i\right)$
[sum of two sides of a traingle is greater than the third side]
In $\Delta OBC,OB+OC>BC.....\left(ii\right)$
[sum of two sides of a triangle is greater than the third side]
In $\Delta OCD,OC+OD>CD...\left(iii\right)$
[sum of two sides of a triangle is greater than the third side]
In $\Delta ODA,OD+OA>DA...\left(iv\right)$
[sum of two sides of a triangle is greater than the third side]
On adding Eqs. (i), (ii), (iii) and (iv), we get
$2\left[\right(OA+OB+OC+OD]>AB+BC+CD+DA\Rightarrow 2[(OA+OC)+(OB+OD)]>AB+BC+CD+DA\Rightarrow 2(AC+BD)>AB+BC+CD+DA[\because OA+OC=ACandOB+OD=BD]\Rightarrow AB+BC+CD+DA<2(BD+AC)$