Answer :

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


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)


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.

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