Show that sum of all sides …
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Sia ? 5 years, 2 months ago
Given: A quadrilateral ABCD
To prove: The sum of all the sides is greater than the sum of its diagonals.
i.e. AB + BC + CD + DA > AC + BD
Construction: Join AC and BD.
Proof: Since sum of any two sides of a triangle is greater than the third side.
Therefore,
In {tex}\triangle{/tex}ABC
AB + BC > AC ...(i)
In {tex}\triangle{/tex}ACD
AD + DC > AC ...(ii)
In {tex}\triangle{/tex}ABD
AB + AD > BD ...(iii)
and in {tex}\triangle{/tex}BCD
BC + CD > BD ...(iv)
adding (i), (ii), (iii) and (iv) we get
2AD + 2DC + 2AB + 2BC > 2AC + 2BD
{tex}\Rightarrow{/tex} AB + BC + CD + DA > AC + BD
Hence proved.
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