ABCD is a quadilatersl Is AB+BC+CD+DA=AC+BD?
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Sia ? 6 years, 3 months ago
Given: A quadrilateral ABCD.
To prove: AB + BC + CD + DA > AC + BD
Proof: In {tex}\Delta ABC{/tex}, we have
AB + BC > AC…(1) [{tex}\because{/tex} Sum of the lengths of any two sides of a triangle must be greater than the third side]
In {tex}\Delta BCD{/tex}, we have
BC + CD > BD...(2) [Same reason]
In {tex}\Delta CDA{/tex}, we have
CD + DA > AC…(3) [Same reason]
In {tex}\Delta DAB{/tex}, we have
AD + AB > BD…(4) [Same reason]
Adding (1), (2), (3) and (4), we get
AB + BC + BC + CD + CD + DA + AD + AB > AC + BD + AC + BD
{tex}\Rightarrow{/tex} 2AB + 2BC + 2CD + 2DA > 2AC + 2BD
{tex}\Rightarrow{/tex} 2(AB + BC + CD + DA) > 2(AC + BD)
{tex}\Rightarrow{/tex} AB + BC + CD + DA > AC + BD
Hence, proved.
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