## Solve Lesson 80 Page 108 Math 10 SBT – Kite>

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)