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Quadrilateral ABCD is convex. Let M be the midpoint of AD, and let N be the midpoint of BC. Prove that MN is greater than or equal to (AB+CD)/2. 

If MN=(AB+CD)/2, then what can you say about quadrilateral ABCD. 

 Jan 19, 2020
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Let P be the midpoint of diagonal BD.  Then by the trangle inequality, MP + PN > MN.  Also, BP + PN > BN and DP + PM > DM.  Then 2DP + 2PN > BC and 2MP + DP > AD.  Subtracting these inequalities, we get 2PN - 2MP > BC - AD.  Putting this together with the inequality MP + PN > MN gives you MN < (AB + CD)/2.

 

If MN = (AB + CD)/2, then everything must be parallel, so ABCD is a rectangle.

 Jan 23, 2020

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