Question

State true or false:

In a quadrilateral $ABCD$, $AB+BC+CD+DA>2BD$

• A. True
• B. False

Answer 1

The correct option is A True

In order to prove that the sum of the lengths of the diagonals is less than the perimeter, we need to see that the diagonal BD divides the quadrilateral ABCD into two triangles.

Using the triangle inequality:

$|BD|<|AB|+|BC|$ and $|BD|<|CD|+|AD|$

So, $2\times |BD|<|AB|+|BC|+|CD|+|AD$|

So, the statement is true.