From a point inside a quadrilateral (not on any diagonal) we join the vertices of the quadrilateral let’s prove that the sum of length of these line segment is greater than the sum of lengths of the diagonals.

Now let’s see for which position of the point inside the quadrilateral the sum of the lengths of the line segment obtained by the vertices of the quadrilateral with the point will be the smallest.

In the figure shown below, ABCD is a quadrilateral and AC and BD are the two diagonals of quadrilateral.

Also,

E is a point inside the quadrilateral,

We need to prove that,

AE + BE + ED + EC > AC + BD

Now, we know that the straight line is the shortest distance between two points. Therefore, AC will be the shortest distance between A and C and thus AC > AE + EC….(1)

Also,

BD > BE + ED….(2)

Adding (1) and (2), we get,

AC + BD > AE + EC + BE + ED

Hence, Proved.

We can see that at the point where both the diagonals meet the lengths of the line segment obtained by the vertices of the quadrilateral with that point will be smallest.