6. Theorem of Ceva, Menelaus and Van Aubel.

Figure 6:
and so result is true for any β and
γ.
Now suppose the lines XY and BC intersect
at a point Z.
Consider the triangle AA1B (Figure
7). Since M, X and Z are collinear,
|Y C| |MA|
.
|Y A| |MA1| .|ZA1|
|ZC|
= 1.
Then β( |XB| C|
) + γ(|Y
|XA| |Y A| )
= β( |MA1||ZB|
) + γ(|MA1||ZC|
|MA||ZA1| |MA||ZA1| )
|MA1|
=
{β|ZB| + γ|ZC|}
|MA||ZA1|
|MA1|
=
|MA||ZA1| {β|ZA1| − β|BA1| + γ|ZA1| + γ|A1C|}
|MA1|
= (β + γ)
|MA||ZA1| .|ZA1|,
since |BA1|
|A1C|
= γ
β ,
= (β + γ) |MA1|
, as required.
|MA|
6