An inequality on the diagonals of a convex quadrilateral

[Derrida]

Hi,

It is well known by Varignon's Theorem and the Triangle Inequality that in a convex quadrilateral ABCD with midpoints a of side AB, b of side BC, ... d of side DA and perimeter p,

\[ AC + BD < p\]


where AC and BD are the diagonals of the quadrilateral. However, how do I obtain the lower bound to this inequality, namely that


\[ \frac{p}{2} < AC + BD \]


I'm not looking for answers to this question, only ideas that help guide me to the solution of this problem. Thanks very much.

Sorry I'm not very familiar with the latex commands in this forum, I tried enclosing [itex] [/itex] around my formulas but something funny appears (Both display just AC + BD < like that).

 

[pat_connell]

One idea is to use the fact that the perimeter of a convex quadrilateral is the sum of its sides. Thus, p = AB + BC + CD + DA. Also, since the midpoints of the sides are known, we can express each side as twice the sum of two of the midpoints. For example, AB = 2(a + b). Combining these two equations gives us p = 2(a + b + c + d). We can then use this equation to substitute for p in the first inequality. This will give us an expression for AC + BD in terms of the midpoints. We can then try to use the Triangle Inequality or other properties of convex quadrilaterals to find the lower bound for AC + BD.