Assume point $D$ is in the Deep End of $\angle ABC$ and $E$ is a point on $\overrightarrow{BD}$. Prove $E$ is in the Deep End of $\angle ABC$.

by HighSchool15   Last Updated April 25, 2018 17:20 PM

Assume point $D$ is in the Deep End of $\angle ABC$ and $E$ is a point on $\overrightarrow{BD}$ such that $B-D-E$. Prove $E$ is in the Deep End of $\angle ABC$.

Definition of the deep end: For an $\angle ABC$, there are points $D$ in the interior angle of $ABC$ so that every line through $D$ cannot cross both rays $BA$ and $BC$. These points $D$ are in the "Deep End" of the interior angle of $ABC$.

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My idea is to suppose that $E$ is not in the deep end. So then the ray containing it will cross both $BA$ and $BC$ which would contradict the assumption that $D$ is in the deep end. I don't feel this is fully justified though.



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