一條數學題不明白!中二

Let AE=x cm , then ED=(10-x) cm
Note that EB=ED=(10-x) cm
In tri. ABE ,
8^2+x^2=(10-x)^2 (pyth. thm)
64+x^2=100-20x+x^2
20x=36
x=9/5
In tri. BC'F , by letting FC'=y cm
Use the above method , we can also obtain y=9/5
[Note that FC'=FC and BC'=DC]
Let Y be a point on EB such that EY⊥YF
In tri. EYF ,
EY=EB-FC'
EY=(10-(9/5))-(9/5)
EY=32/5 cm
by pyth. thm
EY^2+YF^2=EF^2
(32/5)^2+8^2=EF^2 [Note that YF=BC']
EF=√(2624/25)=(8√41)/5