Given ABCD is a quadrilateral.
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Construction:
Join diagonals AC and BD.
In
ΔABC,AB+BC>AC ...(I) [sum of two sides of a triangle is greater than the third side]
In
ΔBCD,BC+CD>BD ...(ii) [sum of two sides of a traingle is greater than the third side]
In
ΔCDA,CD+DA>AC ...(iii) [sum of two sides of a triangle is greater than the third side]
In
ΔDAB,DA+AB>BD ...(iv) [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)
⇒ AB + BC + CD + DA > AC + BD