Question

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

Answer 1

Hint: In this question first we should observe the figure carefully. After that we find the triangles in the figure and the diagonals. Then, using the triangle inequality formula, the sum of any two sides of a triangle is greater than the third side we find two different values and then by adding them we should be able to find the solution.

Complete step by step answer:
We have to prove that OA+OB+OC+OD > AC+BD.
So by looking at the figure we can say that,
AC and BD are The diagonals of the quadrilateral ABCD.
Now, consider triangle AOC & BOD , where ‘O’ is any arbitrary point inside the quadrilateral.
By triangle inequality we know that the sum of any two sides of a triangle is always greater than the third side.
Thus,
In $\Delta BOD$
$OB + OD > BD \to (1)$
Similarly, In $\Delta AOC$
$OA + OC > AC \to (2)$
Adding (1) and (2), we obtain
$\therefore OB + OD + OA + OC > AC + BD$

Note:
In this question first one should observe the figure carefully. Then one should be careful about the figures and their names as this can lead to confusion. Then one can make mistakes about the triangle inequality rule. Keep in mind the rule as it is the main concept of the question. By these basics you should be able to solve the question.