Here are two practice problems to test that you understand the proofs of Ceva and Menelaus.

Question 1.

In the barycentric coordinates relative to $ \triangle ABC $ on which cevians does the point $ G = \frac{2A + 3B + 5C}{10} $ lie? $

$
**Hint**: Rewrite the expression for $ G $ so that
$ G = xC + yC' $ where $ C' $ lies on the opposite
side from $ C $ .

Answer. [show] ▶

In effect, you need to find $ C' $ for which $ C'(AB)$ and $ G(C C') $, for all three points . Here it is for $ C' $ . Rewrite $ G = ( \frac{2A+3B}{5} ) \frac{5}{10} + \frac{5}{10}C $ which has the form $ G = C' \frac{5}{10} + \frac{5}{10}C $. Therefore $ C' = (\frac{2A+3B}{5} ) $ is the point you seek.

Question 2.

In the barycentric coordinates relative to $ \triangle ABC $ let the coordinates be called $ (a,b,c) $ for $ aA+bB+cC $. Where does the line $ 2a + 3b + 5c =0 $ cross the sides of the triangle? $

$.
**Hint**: We have that $ a=0 $ for all points on $ (BC) $.
And we solve $ 3b+5c=0 $ for $ b= -5 $ and $ c=3 $.
But $ (0,-5,3) $ are not the barycentric coordinates of any point
because they do not add to one, $ 0 \ne 0-5+3=-2 . $

Answer. [show] ▶

The point $ A' = \frac{5}{2}B - \frac{3}{2}C $ lies on $ (BC) $ and on the line $ 2a+3b+5c=0 $ and is therefore 1/3 of our answer.

As in Cartesian coordinates systems, the coordinates in barycentric coordinate stystems are unique, just as long as they add up to $ 1 $. We can use this uniqueness to solve many geometrical problems.

Corrected 10sep09 by WYS