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.

 

[jvicens]

Hello,

Thank you for bringing up this interesting question. It is true that Varignon's Theorem and the Triangle Inequality can be used to prove that $AC + BD < p$ in a convex quadrilateral. However, to obtain the lower bound of $\frac{p}{2} < AC + BD$, we can use a different approach.

First, let's consider the case where the quadrilateral ABCD is a square. In this case, the diagonals are equal in length and also equal to the side length of the square. Therefore, we have $AC + BD = 2 \times \frac{p}{4} = \frac{p}{2}$, which satisfies the lower bound we are looking for.

Now, let's consider a general convex quadrilateral ABCD. We can divide the quadrilateral into two triangles by drawing a diagonal, say AC. By the Triangle Inequality, we know that $AC < AB + BC$. Similarly, $AC < AD + DC$. Therefore, we can write $AC < \frac{p}{2}$ since the sum of all sides of the quadrilateral is equal to the perimeter $p$. Similarly, we can show that $BD < \frac{p}{2}$.

Combining these two inequalities, we get $AC + BD < \frac{p}{2} + \frac{p}{2} = p$. This shows that the lower bound of $\frac{p}{2}$ is satisfied for the diagonals of any convex quadrilateral ABCD.

I hope this helps guide you towards the solution of this problem. Let me know if you have any further questions or if you need any clarification.