# Crossbar Theorem

## Theorem

Let $\triangle ABC$ be a triangle.

Let $D$ be a point in the interior of $\triangle ABC$.

Then there exists a point $E$ such that $E$ lies on both $AD$ and $BC$.

## Proof

Aiming for a contradiction, suppose $BC$ does not meet ray $\overrightarrow {AD}$.

Either $BC$ meets line $\overleftrightarrow {AD}$ or it does not.

If it meets $\overleftrightarrow {AD}$, by the Line Separation Property it must meet the ray opposite to $\overrightarrow {AD}$ at a point $E \ne A$.

According to Proposition 3.8 (b), $E$ is not in the interior of $\angle CAB$.

Point $B$ does not lie on $\overleftrightarrow {AD}$; this is because $D \in \operatorname {int} \angle CAB$, so $D$ and $C$ lie on the same side of $\overleftrightarrow {AB}$, so $D \notin \set {\overleftrightarrow {AB} }$, so $B \notin \set {\overleftrightarrow {AB} }$.

Thus, since $E \in \set {\overleftrightarrow {AD} }$, we have $E \ne B$.

By the same reasoning with $C$ and $B$ interchanged, we have $E \ne C$.

Since $E \in BC$ and $E$ is not an endpoint, we have $B ∗ E ∗ C$.

Thus by Proposition 3.7, $E$ is in the interior of $\angle CAB$, which is a contradiction.

Thus $\overleftrightarrow {AD}$ does not meet $BC$ at all; that is, $B$ and $C$ are on the same side of $\overleftrightarrow {AD}$.

By B-2, we have a point $E$ such that $C ∗ A ∗ E$.

By Lemma 3.2.2, $C$ and $E$ are on opposite sides of $\overleftrightarrow {AD}$.

Thus, by B-4(iii), E and B are on opposite sides of $\overleftrightarrow {AD}$.

But by Proposition 3.8(c), $B$ is on the interior of $\angle DAE$, so $E$ and $B$ are on the same side of $\overleftrightarrow {AD}$

This is a contradiction.

Thus $\overrightarrow {AD}$ meets $BC$.