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Show that in a quadrilateral ABCD, AB+BC+CA+DA > AC+BD.

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take two triangles ABC and triangles BDC

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Other answers:

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

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