Exterior angle bisector

In the following figure $\mathrm{AE}$ is the exterior angle bisector of the angle $\mathrm{A}$ of the triangle $\mathrm{AB} \Gamma$ .

Rendered by QuickLaTeX.com

Prove that $\displaystyle \overrightarrow{\mathrm{AE}} = \frac{1}{\beta-\gamma} \left ( \beta \overrightarrow{\mathrm{AB}} -\gamma \overrightarrow{\mathrm{A} \Gamma} \right )$ .

Solution

It follows from the exterior angle bisector theorem that

(1) $\begin{equation*} \overrightarrow{\Gamma \mathrm{E}} = \frac{\beta}{\gamma} \cdot \overrightarrow{\mathrm{BE}} \end{equation*}$

Hence,

$\begin{align*} \overrightarrow{\mathrm{BE}} - \overrightarrow{\Gamma \mathrm{E}} = \overrightarrow{\mathrm{B} \Gamma} = \overrightarrow{\mathrm{A} \Gamma} - \overrightarrow{\mathrm{AB}} &\overset{(1)}{\Leftrightarrow} \frac{\gamma - \beta}{\gamma} \cdot \overrightarrow{\mathrm{BE}} = \overrightarrow{\mathrm{A} \Gamma} - \overrightarrow{\mathrm{AB}} \\ &\Leftrightarrow \frac{\gamma - \beta}{\gamma} \cdot \left ( \overrightarrow{\mathrm{AE}} - \overrightarrow{\mathrm{AB}}\right ) = \overrightarrow{\mathrm{A} \Gamma} - \overrightarrow{\mathrm{AB}} \\ &\Leftrightarrow \frac{\gamma - \beta}{\gamma} \cdot \overrightarrow{\mathrm{AE}} = \left ( \frac{\gamma - \beta}{\gamma} -1 \right ) \cdot \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{A} \Gamma} \\ &\Leftrightarrow \overrightarrow{\mathrm{AE}} = \frac{1}{\beta - \gamma} \left ( \beta \cdot \overrightarrow{\mathrm{AB}} -\gamma \cdot \overrightarrow{\mathrm{A} \Gamma} \right ) \end{align*}$

Exterior angle bisector

They might interest you:

A beautiful Gamma series

An arithmotheoretic sum

Range of convergence

A logarithmic integral

Gamma inequality

Floor series

Pell-Lucas series

On the geometrical view of an integral

No such function

Leave a Reply Cancel reply