A teacher asked me to prove the well known Crossbar theorem. I tried it in the following way:-
Given: If $D$ is in the interior of $\triangle ABC$, then prove that $\overrightarrow{AD}$ intersects $\overline{BC}$.
Proof: Take $F-A-C$.
$1.$ $F$ and $C$ are on the opposite side of $\overleftrightarrow{AB}$. ($D$ is in the interior of $\triangle ABC$)
$2.$ $D$ is on $C$ side of $\overleftrightarrow{AB}$.
$3.$ $F$ and $D$ are on opposite side of $\overleftrightarrow{AB}$.
$4.$ Open ray $\overrightarrow{AD}$ belongs to $D$ side of $\overleftrightarrow {AB}$.
$5.$ $\overleftrightarrow{AD}$ doesn't intersect $\overline{FB}$ for $A \not=B$.
$6.$ $\overrightarrow{AG}$ and $\overrightarrow{AD}$ are opposite open rays. ($D-A-G$)
$7.$ $\overleftrightarrow{AC}$ separates plane of $\triangle FBC$ into two open (opposite) half-planes, one belonging to $B$ side of open ray $\overrightarrow{FB}$ and other belonging to $G$ side of open ray $\overrightarrow{AG}$.
$(A)$Open ray $\overrightarrow{AG}$ doesn't intersect $\overline {FB}$ for $A\not=F.$
$(B)$$\overline {FB}$ doesn't contain $A$ and $\overrightarrow{AG}$ doesn't contain $F.$
$8.$ Given $\overleftrightarrow{AD}=\overrightarrow{AD}+\overrightarrow{AG}$ deosn't intersect $\overline{FB}$.
$9.$ $B$ is on $F$ side of $\overleftrightarrow{AD}$.
$10.$ $C$ and $F$ are on opposite side of $\overleftrightarrow{AD}$.
$11.$ This implies that $B$ and $C$ are on opposite side of $\overleftrightarrow{AD}$.
$12.$ Hence proved that $\overleftrightarrow{AD}$ intersects $\overline{BC}$.
I was told that there is something wrong with this proof, but I can't find what… This seems fine to me. Can anyone point me out the mistake?
Best Answer
You had to prove that ray $\overrightarrow{AD}$ intersects $\overline{BC}$; however, you proved only that line $\overleftrightarrow{AD}$ intersects $\overline{BC}$. The missing final steps are:
$13.$ $\overrightarrow{AG}$ and $\overline{BC}$, except for point $C$, lie on the opposite sides of $\overleftrightarrow{AC}$, so they have no points in common.
$14.$ That implies $\overrightarrow{AD}$ intersects $\overline{BC}$.
Also, there is a typo in step $5$: it should read $\overrightarrow{AD}$ instead of $\overleftrightarrow{AD}$.