Suppose that points A, B, C form a triangle and that A' $\in$ $l_{BC}$ and B' $\in$ $l_{AC}$ and C' $\in$ $l_{AB}$ are such that the lines $l_{AA'}$, $l_{BB'}$, $l_{CC'}$ are concurrent at G.
Compute: $\frac{A-G}{A-A'}+\frac{B-G}{B-B'}+\frac{C-G}{C-C'}$
I tried expanding the fractions by multiplying by $(A-A')(B-B')(C-C')$ but that gave me a terribly large expression.
I know that if they are concurrent with A', B', C' being the midpoints on the opposite sides, then G will be 2/3 of the distance from A to A', so then I would expect $\frac{A-G}{A-A'}=\frac{\frac{2L}{3}}{\frac{3L}{3}}$ where L is the length from A to A' and the total expression summing to 2. Is this correct? And is this the only option for them all to be concurrent.