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

3 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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  3. Join AD and consider the isosceles triangle ACD. Drop a perpendicular from C to AD and denote it as P
    The points A,C,E and P are concyclic since m(AEC) = m(APC) = 90 degrees

    Join CB and let m(BAC)=X,m(BAD)=Y
    => m(DAC)=m(CDA)=X+Y
    From angles in the same segment we have m(BDC)=X, m(DCB)=Y and m(CBA)=X+Y

    Since ACEP is concylic, => m(EPC)=m(EAC)=m(BAC)=X and m(PCE)=m(PAE)=m(DAE)=Y

    So the triangles APC and BEC are similar (AAA)
    => BE = AP.EC/PC ------------(1)

    Similarly, triangle PEC and DBC are similar
    => BD = EP.CD/PC = PE.AC/PC -----------(2) (since CD=AC)

    (1)+(2) => BE+BD = (AP.EC+PE.AC)/PC -----------(3)

    Applying ptolmey's to the cyclic quadrilateral ACEP
    => AE.PC = AP.EC+AC.PE
    => AE = (AP.EC+PE.AC)/PC -----------(4)

    From (3) and (4), BE+BD=AE Q.E.D

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