Topic
For triangle ABC there are three medians AD, BEIGE, CF. Prove that:
\(\overrightarrow {AD} .\overrightarrow {BC} + \overrightarrow {BE} .\overrightarrow {CA} + \overrightarrow {CF} .\overrightarrow {AB} = 0\)
Solution method – See details
Use the midpoint property of the line segment to transform the left side of the equation
Detailed explanation + Do EASY is the midpoint
BC so \(\overrightarrow {AD} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\) + Do E is the midpoint
AC so \(\overrightarrow {BE} = \frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BC} } \right)\) + Do F is the midpoint
AB
so \(\overrightarrow {CF} = \frac{1}{2}\left( {\overrightarrow {CA} + \overrightarrow {CB} } \right)\)
We have: \(\overrightarrow {AD} .\overrightarrow {BC} + \overrightarrow {BE} .\overrightarrow {CA} + \overrightarrow {CF} .\overrightarrow {AB} \)\( = \frac{1} {2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right).\overrightarrow {BC} + \frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow { BC} } \right).\overrightarrow {CA} + \frac{1}{2}\left( {\overrightarrow {CA} + \overrightarrow {CB} } \right).\overrightarrow {AB} \)
\( = \frac{1}{2}\left( {\overrightarrow {AB} .\overrightarrow {BC} + \overrightarrow {AC} .\overrightarrow {BC} + \overrightarrow {BA} .\overrightarrow {CA}) + \overrightarrow {BC} .\overrightarrow {CA} + \overrightarrow {CA} .\overrightarrow {AB} + \overrightarrow {CB} .\overrightarrow {AB} } \right)\)\( = \frac{1}{2}\left( { – \overrightarrow {BA} .\overrightarrow {BC} + \overrightarrow {CA} .\overrightarrow {CB} + \overrightarrow {AB} .\overrightarrow {AC) } – \overrightarrow {CB} .\overrightarrow {CA} – \overrightarrow {AC} .\overrightarrow {AB} + \overrightarrow {BC} .\overrightarrow {BA} } \right)\)\( = \frac{1 }{2}.0 = 0\) (PPC)
Source link net do edu
