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
May 31, 2018

AB=15AB=15 units

Explanation:

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Let |ABC||ABC| denote area of ABCABC
let |ABC|=12a|ABC|=12a,
as AD=DC, => |BDC|=|BDA|=(12a)/2=6aAD=DC,|BDC|=|BDA|=12a2=6a
as BE=ED, => |BEC|=|DEC|=(6a)/2=3aBE=ED,|BEC|=|DEC|=6a2=3a,
let |BFE|=x, => |DFE|=x|BFE|=x,|DFE|=x,
=> |AFD|=|BDA|-|BDF|=6a-2x|AFD|=|BDA||BDF|=6a2x
As AD=DC, => |AFD|=|CFD|AD=DC,|AFD|=|CFD|,
=> 6a-2x=3a+x6a2x=3a+x
=> color(red)(x=a)x=a
=> |CFB|=3a+x=4a|CFB|=3a+x=4a
=> AB:FB=|ABC|:|CFB|=12a:4a=3:1AB:FB=|ABC|:|CFB|=12a:4a=3:1
=> AB=3*FB=3xx5=15AB=3FB=3×5=15 units