Question

Question 12
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD.

Answer 1

Given ABCD is a quadrilateral.

Construction:
Join diagonals AC and BD.

In $\Delta ABC,AB+BC>AC...\left(I\right)$
[sum of two sides of a triangle is greater than the third side]
In $\Delta BCD,BC+CD>BD...\left(ii\right)$
[sum of two sides of a traingle is greater than the third side]
In $\Delta CDA,CD+DA>AC...\left(iii\right)$
[sum of two sides of a triangle is greater than the third side]
In $\Delta DAB,DA+AB>BD...\left(iv\right)$
[sum of two sides of a triangle is greater than the third side]
On adding Eqs, (i), (ii), (iii) and (iv), we get
2 (AB + BC + CD + DA) > 2 (AC + BD)
$\Rightarrow$ AB + BC + CD + DA > AC + BD