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

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

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,

[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.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top