Answer:
[tex]BN=4\ \text{in}[/tex]
[tex]AN=2\ \text{in}[/tex]
[tex]AC=\sqrt{5}\ \text{in}[/tex]
Step-by-step explanation:
Let [tex]BN=x[/tex]
[tex]BC=c[/tex]
[tex]NC=y=1\ \text{in}[/tex]
[tex]AB=a=2\sqrt{5}\ \text{in}[/tex]
[tex]AC=b[/tex]
We have the relation
[tex]\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5[/tex]
[tex]\boldsymbol{BN=4\ \text{in}}[/tex]
[tex]h=\sqrt{a^2-x^2}\\\Rightarrow h=\sqrt{(2\sqrt{5})^2-4^2}\\\Rightarrow h=2\ \text{in}[/tex]
[tex]\boldsymbol{AN=2\ \text{in}}[/tex]
[tex]b=\sqrt{h^2+y^2}\\\Rightarrow b=\sqrt{2^2+1^2}\\\Rightarrow b=\sqrt{5}\ \text{in}[/tex]
[tex]\boldsymbol{AC=\sqrt{5}\ \text{in}}[/tex]
