Prove
that in any quadrilateral, the sum of the diagonals is greater than
the half of its perimeter.
Consider
the quad. In the above diagram. Let E be the point of the
intersection.
Now, AE+EB
> AB
EB+EC
> BC
AE+ED
> AD
EC+ED
> DC (Using Triangle Inequality)
Adding
the above four inequalities we get
2(AE+EC+EB+ED)
> AB+BC+AD+DC
=>
AC + BD > ½(AB+BC+AD+DC)
Thus
sum of the diagonals is greater than the half of its perimeter Q.E.D
In
any triangle four times the sum of its medians is greater than 3
times its perimeter.
We
know that difference of any two sides of a triangle is less than the
third side (prove it)
In
triangle ABE,
AE
> AB-BE
In
triangle ACE,
AE
> AC-CE
Adding
above two inequalities we get,
2AE
> AB + AC -(BE+CE)
=>
AE > ½(AB+AC-BC)
=>
4AE > 2(AB+AC-BC).........(1)
Similarly,
4BD
> 2(AB+BC-AC).............(2) and 4CF > 2(AC+BC-AB)........(3)
Adding
(1),(2) and (3) we have,
4(AE+BD+CF)
> 2(AB+AC-BC+AB+BC-AC+AC+BC-AB)
=>
4(AE+BD+CF) > 2(AC+AB+BC)
=>
sum of the lengths of the medians is greater than half the perimeter
We
can strengthen the inequality by using the fact that the point 'O'
divides the medians AE,BD,CF internally in the ration 2:1
Therefore,
OD:OB = 1:2
=>
(OB+OD):OB = (1+2):2
=>
BD:OB=3:2
=>
OB = 2/3 BD........(a)
Similarly,
OC = 2/3 CF.........(b) and OA = 2/3 AE.......(c)
Now
in triangle OBC, OB+OC> BC
=>2/3(BD+CF)>BC
[using (a) and (b)]
=>
2(BD+CF)>3BC
Similarly,
2(CF+AE)>3AC and 2(BD+AE)>3AB
Adding
the last three inequalities we get 4(AE+BD+CF) > 3(AB+BC+CA)
In the triangle ABC, AE,BD and CF are the medians where O is the point of there intersection
