An inequality on the diagonals of a convex quadrilateral (1 Viewer)

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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,

[itex] AC + BD < p [/itex]

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

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

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

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