# Let’s prove that the perimeter of a quadrilateral is more than twice of the sum length of any diagonal.

Let us first draw a figure for above question. AC is one of the diagonals.

The diagonal AC divides the quadrilateral into two triangles.

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

Let us first consider Δ ACD

AD + DC > AC ……….. (i)

Similarly let us consider the other Δ ABC

AB + BC > AC ………… (ii)

Now let us add the equations (i) and (ii)

AD + AC + BC + AB > AC + AC

AD + AC + BC + AB > 2 × AC

Hence it is proved that the perimeter of a quadrilateral is more than twice of the sum length of any diagonal.

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