Orthic Triangle of Obtuse Triangle
Jump to navigation
Jump to search
Theorem
Let $\triangle ABC$ be an obtuse triangle such that $A$ is the obtuse angle.
Let $H$ be the orthocenter of $\triangle ABC$.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.
Then $\triangle DEF$ is also the orthic triangle of $\triangle HBC$, which is an acute triangle.
Proof
By construction:
- $CE \perp BH$
- $BF \perp CH$
- $HD \perp BC$
Thus by definition $CE$, $BF$ and $HD$ are the altitudes of $\triangle HBC$.
Also by construction, $A$ lies on $CE$, $BF$ and $HD$.
Hence $\triangle DEF$ is the orthic triangle of $\triangle HBC$.
This needs considerable tedious hard slog to complete it. In particular: Demonstrate that $\triangle HBC$ is acute To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |