In convex quadrilateral ABCD, K is midpoint of AB, M is midpoint CD,. Prove that $ KM \le \frac{BC+AD}{2} $.

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- Oct 26th 2010, 09:32 AMolkaabcinequality in quadrilateral
In convex quadrilateral ABCD, K is midpoint of AB, M is midpoint CD,. Prove that $ KM \le \frac{BC+AD}{2} $.

- Oct 26th 2010, 10:06 AMWilmer
- Oct 26th 2010, 10:35 AMPlato
We know that $\displaystyle \frac{1}{2}\overrightarrow {AB} + \overrightarrow {BC} + \frac{1}{2}\overrightarrow {CD} = \overrightarrow {KM} $ and $\displaystyle -\frac{1}{2}\overrightarrow {AB} + \overrightarrow {AD} - \frac{1}{2}\overrightarrow {CD} = \overrightarrow {KM} $

If you add those two together and divide by two, what do get?