Benutzer:MovGP0/Dreieck

Given by the Sinussentence:

${\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}=2r$

where r is the radius of the Outher-Circle; and $l_{a}={\frac {\alpha }{2}}$

From these follows:

$\operatorname {\angle } \left({\overline {AL_{a}}},{\overline {BL_{a}}}\right)=\arcsin \left({\frac {\overline {AB}}{\overline {BL_{a}}}}\sin l_{a}\right)$	$\operatorname {\angle } \left({\overline {AL_{a}}},{\overline {CL_{a}}}\right)=\arcsin \left({\frac {\overline {AC}}{\overline {CL_{a}}}}\sin l_{a}\right)$
${\overline {BL_{a}}}={\overline {AL_{a}}}{\frac {\sin l_{a}}{\sin \beta }}$	${\overline {CL_{a}}}={\overline {AL_{a}}}{\frac {\sin l_{a}}{\sin \gamma }}$
${\overline {AD}}={\overline {AB}}{\frac {\sin \beta }{\sin \operatorname {\angle } \left({\overline {AL_{a}}},{\overline {BL_{a}}}\right)}}$	${\overline {AD}}={\overline {AC}}{\frac {\sin \gamma }{\sin \operatorname {\angle } \left({\overline {AL_{a}}},{\overline {CL_{a}}}\right)}}$
$\beta =\pi -l_{a}-\operatorname {\angle } \left({\overline {AL_{a}}},{\overline {BL_{a}}}\right)$	$\gamma =\pi -l_{a}-\operatorname {\angle } \left({\overline {AL_{a}}},{\overline {CL_{a}}}\right)$

Solving to ${\overline {AD}}$ gives:

${\overline {AL_{a}}}={\frac {\sin \left[{\pi -l_{a}-\arcsin \left({\frac {\overline {AB}}{\overline {BL_{a}}}}\sin l_{a}\right)}\right]}{\sin l_{a}}}={\frac {\sin \left[{\pi -l_{a}-\arcsin \left({{\frac {\overline {AC}}{\overline {CL_{a}}}}\sin l_{a}}\right)}\right]}{\sin l_{a}}}$

witch can solved to the expression:

${\frac {\overline {AB}}{\overline {BL_{a}}}}={\frac {\overline {AC}}{\overline {CL_{a}}}}$

Witch in fact shows us that each bisector in the triangle divides the side opposite in the relationship of the sides lying close.

Apply Cosine

It follows from the projection of the generalized sentence of Pythagoras to cosine $c^{2}=a^{2}+b^{2}+2ab\cos \gamma$:

${\overline {AL_{a}}}^{2}={\overline {AB}}^{2}+{\overline {BL_{a}}}^{2}+2\cdot {\overline {AB}}\cdot {\overline {BL_{a}}}\cdot {\frac {AL_{a}}{BL_{a}}}={\overline {AC}}^{2}+{\overline {CL_{a}}}^{2}+2\cdot {\overline {AC}}\cdot {\overline {CL_{a}}}\cdot {\frac {AL_{a}}{CL_{a}}}$ (the cosine already solved)

gives:

${\overline {ALa}}={\frac {1}{2}}\cdot {\frac {{\overline {AB}}^{2}+{\overline {BL_{a}}}^{2}-\left({\overline {AC}}^{2}+{\overline {CL_{a}}}^{2}\right)}{{\overline {AC}}-{\overline {AB}}}}$

bring it to end

TODO

Result

${{\overline {AL_{a}}}\cdot {\overline {AL_{a}}}+{\overline {BL_{a}}}\cdot {\overline {CL_{a}}}={\overline {AB}}\cdot {\overline {AC}}}$

Solved to ${\overline {AL_{a}}}$:

${{\overline {AL_{a}}}={\sqrt {{\overline {AB}}\cdot {\overline {AC}}-{\overline {BL_{a}}}\cdot {\overline {CL_{a}}}}}}$