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