In triangle ABC, BD is a median. CF intersects BD at E so that BE = ED. Point F is on AB. Then, if BF = 5. What is the value of AB?

1 Answer
CW
May 31, 2018

#AB=15# units

Explanation:

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Let #|ABC|# denote area of #ABC#
let #|ABC|=12a#,
as #AD=DC, => |BDC|=|BDA|=(12a)/2=6a#
as #BE=ED, => |BEC|=|DEC|=(6a)/2=3a#,
let #|BFE|=x, => |DFE|=x#,
#=> |AFD|=|BDA|-|BDF|=6a-2x#
As #AD=DC, => |AFD|=|CFD|#,
#=> 6a-2x=3a+x#
#=> color(red)(x=a)#
#=> |CFB|=3a+x=4a#
#=> AB:FB=|ABC|:|CFB|=12a:4a=3:1#
#=> AB=3*FB=3xx5=15# units

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