• October 26th 2010, 09:32 AM
olkaabc
In convex quadrilateral ABCD, K is midpoint of AB, M is midpoint CD,. Prove that $KM \le \frac{BC+AD}{2}$.
• October 26th 2010, 10:06 AM
Wilmer
Quote:

Originally Posted by olkaabc
Prove that $KM \le \frac{BC+AD}{2}$.

Huh?
• October 26th 2010, 10:35 AM
Plato
Quote:

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

We know that $\frac{1}{2}\overrightarrow {AB} + \overrightarrow {BC} + \frac{1}{2}\overrightarrow {CD} = \overrightarrow {KM}$ and $-\frac{1}{2}\overrightarrow {AB} + \overrightarrow {AD} - \frac{1}{2}\overrightarrow {CD} = \overrightarrow {KM}$

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