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Question Number 28240 by $@ty@m last updated on 22/Jan/18

Answered by ajfour last updated on 22/Jan/18

DE×EA=CE×EF  6×2=6(√2)×EF  ⇒  EF=(√2)  so  CF=7(√2)  let foot of ⊥ from F on BC be G.  CG=7 , GB=1, FG=7  BF^(  2)  =GB^( 2) +FG^( 2)             =1+49 =50  BF =5(√2)  ≈ 7.07

DE×EA=CE×EF6×2=62×EFEF=2soCF=72letfootoffromFonBCbeG.CG=7,GB=1,FG=7BF2=GB2+FG2=1+49=50BF=527.07

Commented by $@ty@m last updated on 22/Jan/18

Thank you very much.

Thankyouverymuch.

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