Answer: Consider a quadrilateral ABCD such that AC ans BD are the two diagonals of ABCD.
With reference to figure 1 .
In triangle ABC and triangle ADC ,by using triangle inequality we have
AB +BC > AC .......(1)
and AD + CD >AC......(2)
adding (1) and (2) we have
AB+ BC+CD+AD>2AC....(3)
Similarly we get for diagonal BD
AB+BC+CD+AD>2BD....(4)
Adding (3) and (4) we get
2(AB+BC+CD+AD)> 2(AC+BD)
⇒(AB+BC+CD+AD)> AC+BD where (AB+BC+CD+AD) = perimeter of quadrilateral.
Hence, the sum of the length of the diagonals of a quadrilateral is less than the perimeter.
Now 2 diagonal divides quadrilateral into 4 quadrilaterals
Therefore AO+OD>AD
OD+OC>CD
OC+OB>BC
And OA+OB>AB
Adding all these conclude that
2(AC+BD)>AB+BC+CD+AD
⇒AC+BD>1/2(AB+BC+CD+AD) where (AB+BC+CD+AD) = perimeter of quadrilateral.
Hence,the sum of the length of the diagonals is greater than the half of the perimeter of this quadrilateral.
Now consider a triangle ABC as given in figure (2)
As we now,the sum of two sides of a triangle is greater than twice the median bisecting the third side.
Therefore AB+BC>2BE
Similarly,
AB+AC>2AD
BC+AC>2CF
By adding all these we get ,
2(AB+BC+CA)>2(AD+CF)
⇒AB+BC+CA>AD+CF
Hence, the sum of the lengths of the three medians in a triangle is smaller than the perimeter of the triangle.