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.)

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.

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step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35

im not sure

step-by-step explanation: