Answer:
[tex]BF=45[/tex]
Step-by-step explanation:
In the triangle [tex]ABF[/tex] and triangle[tex]ACE[/tex]
[tex]BF||CE\\ \Rightarrow \angle ABF=\angle ACE\\ \Rightarrow \angle AFB=\angle AEC\\ \angle A=\angle A[/tex]
Hence from [tex]AAA\ \ [/tex][tex]\Delta ABF\cong \Delta ACE[/tex]
[tex]\frac{BC}{CE}=\frac{AB}{AC}\\\frac{BF}{CE}=\frac{AB}{2AB}\\BF=\frac{1}{2}\times CE\\BF=\frac{1}{2}\times 90\\BF=45[/tex]
