prove that sum of all sides of a quadrilateral is less than sum of diagonals ×2
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In a quad. ABCD,
Where AC and BD are diagonals.
Using prop. of triangles that sum of 2 sides is always greater than the 3rd side;
We 4 triangles in the quad which atleast include 2 sides of the quad. and get our conclusion i.e:
AB+BC>AC
BC+CD>BD
CD+AD>AC
AD+AB>BD
Adding all these
2AB+2BC+2CD+2AD>2AC+2BD.
Dividing by 2
AB+BC+CD+AD>AC+BD .
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