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Given any quadrilateral ABCD. Let X be the midpoint of side AB and Y be the midpoint of side CD. How can I prove that XY is not greater than max{AC, BD} ? Intuitively I see it is true in all cases, but don't have a clue how to prove this.

I started using the argument that XY <= AD, considering AD be the longest diagonal. This gives me two triangles ACD and ADB. Not sure whether that will help eventually or not. –

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    $\begingroup$ I solved your problem. If you want to see my solution, show please your attempts. $\endgroup$
    – Michael Rozenberg
    Mar 3, 2020 at 3:23
  • $\begingroup$ @MichaelRozenberg I started using the argument that XY <= AD, considering AD be the longest diagonal. This gives me two triangles ACD and ADB. Not sure whether that will help eventually or not. $\endgroup$
    – Mateo
    Mar 3, 2020 at 3:51

1 Answer 1

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$$XY=|\vec{XY}|=\left|\frac{1}{2}\left(\vec{AC}+\vec{BD}\right)\right|\leq\frac{1}{2}\left(|\vec{AC}|+|\vec{BD}|\right)\leq\max\{AC,BD\}.$$

I used the following reasoning. $$\vec{XY}=\frac{1}{2}\left(\vec{XA}+\vec{AC}+\vec{CY}+\vec{XB}+\vec{BD}+\vec{DY}\right)=$$ $$=\frac{1}{2}\left(\vec{XA}+\vec{XB}+\vec{CY}+\vec{DY}+\vec{AC}+\vec{BD}\right)=$$ $$=\frac{1}{2}\left(\vec{0}+\vec{0}+\vec{AC}+\vec{BD}\right)=\frac{1}{2}\left(\vec{AC}+\vec{BD}\right).$$

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  • $\begingroup$ How did you come up with $|XY| = |\frac{1}{2}(AC+BD)|$? I did not understand that part. $\endgroup$
    – Mateo
    Mar 3, 2020 at 5:43
  • $\begingroup$ @Mateo I added something. See now. $\endgroup$
    – Michael Rozenberg
    Mar 3, 2020 at 5:50
  • $\begingroup$ I understand now. You used the polygon law of vector addition. The proof looks good. I am trying to prove the same argument using triangle inequality. I will post the solution if I become successful. $\endgroup$
    – Mateo
    Mar 3, 2020 at 5:59
  • $\begingroup$ @Mateo I can prove it by the triangle inequality, but it's a very ugly here. $\endgroup$
    – Michael Rozenberg
    Mar 3, 2020 at 6:01

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