anonymous
  • anonymous
Show that in a quadrilateral ABCD, AB+BC+CA+DA > AC+BD.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1344502321128:dw|
anonymous
  • anonymous
|dw:1344502527479:dw|
anonymous
  • anonymous
take two triangles ABC and triangles BDC

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More answers

anonymous
  • anonymous
|dw:1344502571798:dw|In triangle ABC,
anonymous
  • anonymous
I think the question is AB+BC+CA+DA > AC+CD
anonymous
  • anonymous
@pratu043 i think there is a problem with question
anonymous
  • anonymous
AB+BC+CA+DA > AC+BD --- ---
anonymous
  • anonymous
@pratu043 u there
anonymous
  • anonymous
Sorry, it is BD not CD.
anonymous
  • anonymous
help please.
anonymous
  • anonymous
oh i got it
anonymous
  • anonymous
in triangle ABD , AB + DA > BD
anonymous
  • anonymous
The quadrilateral should be like this: |dw:1344503679049:dw|
anonymous
  • anonymous
in triangle ABD , AB + DA > BD right @pratu043
anonymous
  • anonymous
How did you get that @sauravshakya?
anonymous
  • anonymous
In a triangle sum of two sides of a triangle is always greater than the third side right?
anonymous
  • anonymous
@pratu043
anonymous
  • anonymous
Oops, yes I forgot that.
anonymous
  • anonymous
So what next?
anonymous
  • anonymous
Cancel CA and AC since they are same side..... so u will get AB+BC+DA > BD.
anonymous
  • anonymous
right?
anonymous
  • anonymous
and since AB + BC > BD, AB+BC+DA > BD (thus, proved)
anonymous
  • anonymous
got it?
anonymous
  • anonymous
But AB+BC+DA > BD is not what we should prove.
jiteshmeghwal9
  • jiteshmeghwal9
|dw:1344516745619:dw|

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