User talk:Tsukitakemochi/sandbox

Verification for orthocenter equation
Think $$H$$ be the orthocenter. According to table contents, $$H$$ should be given as


 * $$H=\frac{(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})B+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})C}{(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})}.$$

In order to verify if $$H$$ be unmistakably the orthocenter or not, goal can be if $$(\vec{AH},\vec{BC})=0$$ can be said or not. To avoid chaos, let's apply $$m$$ as the denominator for above $$H$$ equation, like


 * $$m\equiv(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA}),$$

then


 * $$mH=(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})B+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})C,$$


 * $$m\vec{AH}=((\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})B+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})C)$$
 * $$-((\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})A+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})A)$$


 * $$=(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})\vec{AB}+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})\vec{AC},$$


 * $$(m\vec{AH},\vec{BC})=(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})(\vec{AB},\vec{BC})+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})(\vec{AC},\vec{BC})$$


 * $$=(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})(\vec{AB},\vec{BC})-(\vec{AB},\vec{AC})(\vec{AB},\vec{BC})(\vec{CA},\vec{CB})$$
 * $$=0.$$

Usually $$m\neq 0$$. So,


 * $$(\vec{AH},\vec{BC})=0.$$