Wednesday, September 30, 2009

Problem 362. Circle, Chord, Perpendicular, Equal chords

Proposed Problem
Click the figure below to see the complete problem 362 about Circle, Chord, Perpendicular, Equal chords.

 Problem 362. Circle, Chord, Perpendicular, Equal chords.
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Complete Problem 362
Level: High School, SAT Prep, College geometry

2 comments:

  1. join C with B, draw CFL ( F on AB, L on circle) that
    FE = EB ( get ang CBE = ang CFB )

    now have to prove AF = DB

    ang CBE = ang ALF (at the same harc AC )
    => ang ALF = ang AFL ( AFL = CFE = CBE )
    => AL = AF

    now have to prove AL = BD

    ang FLD = CBE ( on equal harcs AC=CBD)
    =>ang FLD = ang AFL
    => AB // LD, => AL = BD, or AF = BD

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  2. Extend AB to F such that AC = CF. Then C is the circumcentre of Tr. ADF. Hence < DCF = 2.< DAF and since< DAC =< DCB it follows that BC bisects< DCF
    Hence Tr.s CDB & CFB are congruent and so BD=BF
    But AE=EF hence AE = EB + BD

    Sumith Peiris
    Moratuwa
    Sri Lanka

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