Geometry Inequality

Post by Maths.

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