Prove that the sum of the length of the diagonals of a quadrilateral is less than the perimeter, but greater than the half of the perimeter of this quadrilateral.

and

Prove that the sum of the lengths of the three medians in a triangle is smaller than the perimeter of the triangle. (The statement is true for the altitudes as well.)

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.