Equality of vectors
Posted: Sat Apr 06, 2024 8:50 am
In the following figure $\mathrm{AE}$ is the exterior angle bisector of the angle $\mathrm{A}$ of the triangle $\mathrm{AB} \Gamma$.
Prove that $\displaystyle \overrightarrow{\mathrm{AE}} = \frac{1}{\beta-\gamma} \left ( \beta \overrightarrow{\mathrm{AB}} -\gamma \overrightarrow{\mathrm{A} \Gamma} \right )$.
- quicklatex.com-4a9d5cf71853d480c0d967a5686094ba_l3.png (3.65 KiB) Viewed 1744 times
Prove that $\displaystyle \overrightarrow{\mathrm{AE}} = \frac{1}{\beta-\gamma} \left ( \beta \overrightarrow{\mathrm{AB}} -\gamma \overrightarrow{\mathrm{A} \Gamma} \right )$.