Answer :


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.


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