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Let ABCD be a convex quadrilateral, and let M and N be the midpoints of sides AD and BC, respectively. Prove that MN<=(AB+CD)/2. When does equality occur?

Thanks!

 Dec 16, 2018
 #1
avatar+700 
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Let's just say that ABCD is a trapezoid. M and N are the midpoints of the non-parallel sides AD and BC, respectively. There is a theorem that states the line MN is the average of the two parallel sides (AB and CD). 

 

If you try a quadrilateral (cannot be trapezoid or parallelogram) with an extremely long side, you will find that MN < (AB+CD)/2.

 

This equality works for all convex and real quadrilaterals. 

 

Hope this helps,

- PM

 Dec 16, 2018
 #2
avatar+103689 
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That is not helpful MP.  

The question says 'prove' (the inequality)  and you have not even attempted to prove anything.

 

I suppose you did more or less state when the equality occurs. I can give you credit for that bit.   :)

Melody  Dec 16, 2018
edited by Melody  Dec 16, 2018
 #3
avatar+103689 
+1

I'd like to know how to do this too. :/

 

It is easy to prove the equality for the trapezium but I have not worked out how to prove it for the inequality. 

 Dec 16, 2018
 #4
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GingerAle  Dec 16, 2018
 #5
avatar+103689 
+2

Thanks!

I thought I had done it before but the method wasn't coming to me.    laugh

 

Now I can learn from my own answer LOL 

Melody  Dec 16, 2018

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