Question

If O is a point within a quadrilateral ABCD, show that $OA+OB+OC+OD>AC+BD.$

Answer 1

Given:- $ABCD$ is a quadrilateral. $O$ is a point inside the quadrilateral $ABCD$.

To prove:- $OA+OB+OC+OD>AC+BD$

Construction : Join $OA,OB,OC$ and $OD$. Also, join $AC$ and $BD$.

Proof:- As we know that the sum of any two sides of a triangle is greater than the third side.

Therefore,

In $△BOD$,

$OB+OD>BD.....\left(1\right)$

Similarly

In $△AOC$,

$OA+OC>AC.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$OB+OD+OA+OC>BD+AC$

$\therefore OA+OB+OC+OD>AC+BD$

Hence proved.