Rysunek poglądowy:

 

Z twierdzenia cosinusów w trójkącie ABC:

${\left|AB\right|}^2={\left|BC\right|}^2+{\left|CA\right|}^2-2\cdot \left|BC\right|\cdot \left|CA\right|\cdot \cos \alpha$  

$49=81+100-2\cdot 9\cdot 10\cdot \cos \alpha$  

$49=181-180\cos \alpha$  

$-132=-180\cos \alpha$  

$\cos \alpha =\frac{11}{15}$  

 

Z twierdzenia cosinusów w trójkącie BCD:

${\left|BD\right|}^2={\left|BC\right|}^2+{\left|CD\right|}^2-2\cdot \left|BC\right|\cdot \left|CD\right|\cdot \cos \alpha$  

${\left|BD\right|}^2={\left|BC\right|}^2+{\left(\frac{1}{2}\cdot \left|CA\right|\right)}^2-\left|BC\right|\cdot \left|CA\right|\cdot \cos \alpha$  

${\left|BD\right|}^2=81+25-9\cdot 10\cdot \frac{11}{15}=81+25-90\cdot \frac{11}{15}=106-66=40$  

$\left|BD\right|=\sqrt{40}=\sqrt{4}\cdot \sqrt{10}=2\sqrt{10}$