Prove that the perimeter of any quadrilateral is greater than twice the length of any of its diagonal

Question

I am stuck with the following problem that says :

Prove that the perimeter of any quadrilateral is greater than twice the length of any of its diagonal.

My try: ........

For any quadrilateral $ABCD\,,$ we can easily prove that $$AB+BC+CD+DA \gt AC+BD......\tag{1}$$

Now, three cases arise. Either, $$AC=BD\,\,or AC \gt BD\,\, or AC \lt BD$$.

If $AC=BD$,then the result follows from (1).

If $AC \gt BD \implies AC+BD \gt 2BD $ and then the result follows from (1).

If $ BD \gt AC \implies AC+BD \gt 2AC $ and then the result follows from (1).

Can someone take some time to check if I made any mistake or is there any better way to tackle the problem.

Thanks in advance for your time.

Answer 1

$AB+AD>BD$ and $BC+CD>BD$, then $$AB+BC+CD+DA>2BD$$

Answer 2

Remove one of the diagonals to help the visual. Then you have two triangles. Apply triangle inequality.

Answer 3

It is just a straightforward implementation of the properties of a triangle. $$AB + BC > AC$$ $$ AD + CD > AC $$

Similarly for the other diagonal.

Answer 4

We assume the original quadrilateral is nondegenerate (not a point or a line segment).

Pick a diagonal $D_1$, and consider the family of quadrilaterals obtained by moving the other two vertices linearly along the other diagonal $D_2$ toward the center. As the vertices move in, the perimeter of the resulting quadrilaterals strictly shrinks.

Initially, the perimeter is that of the original quadrilateral; in the limiting case, the vertices meet in the center and the perimeter has shrunk to exactly twice the length of $D_1$.

So the original perimeter is strictly less than twice the length of any given diagonal.

Answer 5

E be the meeting point of diagonals... ∆AEB , AE+EB > AB ∆EBC , EC + EB >BC ∆ CED, CE + DE > CD ∆ DEA, DE + EA > DA Adding all above 2AC+ 2BD> AB+BC+CD+DA(Replacing BE and ED with BD, similarly with AC)