From One Collinearity to Another

The following theorem is due to Dao Thanh Oai (this is theorem 5 from his online paper):

Let collinear points $A_1,$ $B_1,$ $C_1$ lie on the sidelines $BC,$ $AC,$ and $AB$ of $\Delta ABC.$ Assume points $A_2,$ $B_2,$ $C_2$ are collinear with $A_1,$ $B_1,$ $C_1$ and define $A_3=BC\cap AA_2,$ $B_3=AC\cap BB_2,$ $C_3=AB\cap CC_2.$

Assume also that

$\displaystyle\frac{A_2C_1}{A_2B_1}\cdot\frac{B_2A_1}{B_2C_1}\cdot\frac{C_2B_1}{C_2A_1}=-1.$

Then points $A_3,$ $B_3,$ $C_3$ are collinear.

Note that the required condition is satisfied when $A_2,$ $B_2,$ $C_2$ are the midpoints of $B_1C_1,$ $A_1C_1,$ and $A_1B_1,$ respectively.

Proof

Let's agree that below all the segments are directed.

In $\Delta B_1A_1C$ with a transversal $A_2AA_3,$ the Menelaus theorem gives

$\displaystyle\frac{A_2B_1}{A_2A_1}\cdot\frac{A_3A_1}{A_3C}\cdot\frac{AC}{AB_1}=1.$

In $\Delta C_1BA_1$ with a transversal $AA_3A_2,$ we have

$\displaystyle\frac{A_2A_1}{A_2C_1}\cdot\frac{A_3B}{A_3A_1}\cdot\frac{AC_1}{AB}=1.$

The latter two combine to prove

$\displaystyle\frac{A_3B}{A_3C}=\frac{A_2C_1}{A_2B_1}\cdot\frac{AB_1}{AC}\cdot\frac{AB}{AC_1}.$

Similarly we obtain

$\displaystyle\frac{B_3C}{B_3A}=\frac{B_2A_1}{B_2C_1}\cdot\frac{BC_1}{BA}\cdot\frac{BC}{BA_1}.$

and

$\displaystyle\frac{C_3A}{C_3B}=\frac{C_2B_1}{C_2A_1}\cdot\frac{CA_1}{CB}\cdot\frac{CA}{CB_1}.$

Multiplying the three and taking into account the statement of Menelaus' theorem for $\Delta ABC$ and the transversal $C_1A_1B_1,$ i.e.,

$\displaystyle\frac{A_1B}{A_1C}\cdot\frac{B_1C}{B_1A}\cdot\frac{C_1A}{C_1B}=1,$

we get after simplification,

$\displaystyle\frac{A_3B}{A_3C}\cdot\frac{B_3C}{B_3A}\cdot\frac{C_3A}{C_3B}=-\frac{A_2C_1}{A_2B_1}\cdot\frac{B_2A_1}{B_2C_1}\cdot\frac{C_2B_1}{C_2A_1}=-(-1)=1.$

The converse of Menelaus' theorem shows that points $A_3,$ $B_3,$ $C_3$ are indeed collinear.

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