Tuesday, January 25, 2011

Problem 574: Circle, Chord, Secant, Angle Bisector, Perpendicular

Geometry Problem
Click the figure below to see the complete problem 574.

Problem 574: Circle, Chord, Secant, Angle Bisector, Perpendicular.
Zoom at: Geometry Problem 574

Posted by Antonio Gutierrez at 2:40 PM
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9 comments:

  1. PravinJanuary 25, 2011 at 6:13 PM

    Let FM intersection CD be G
    Denote the "acute" angles at
    A, B, C, D, E, F, G
    by the corresponding small letters
    a, b, c, d, e, f, g.
    b = d = f + c
    f = b – c
    d = g + (f/2) ,
    g = d – (f/2) = b – [(b-c)/2] = (b + c)/2
    e + b + c = Pi,
    (e/2) = (Pi/2) – [(b+c)/2] = (Pi/2) – g
    (e/2) + g = Pi/2 in triangle EHG
    So angle at H = 90 degrees

    ReplyDelete
  2. UnknownJanuary 26, 2011 at 1:07 AM

    http://img137.imageshack.us/img137/1631/problem574.png

    EH cut AD at K and BC at L ( see attached picture)
    Note that Triangle KED similar to tri. LEB (case AA)
    So angle(LKF)=angle(KLF) and triangle KLF is isosceles
    And angle bisector FH of isosceles triangle KLF perpendicular to the base KL at H

    Peter Tran

    ReplyDelete
  3. c .t . e. oJanuary 26, 2011 at 1:42 AM

    name P intersection of AC, DB extended
    ▲AEP ~ ▲DEP ( CAB = CDB, AEP = DEP )
    => APH = DPH => PH bisector from 572
    PH perpendicular to FH

    ReplyDelete
  4. Antonio GutierrezJanuary 26, 2011 at 5:34 AM

    Problem #574. To c.t.e.o: What is the reason for your statement angle AEP = angle DEP?

    ReplyDelete
  5. c .t . e. oJanuary 26, 2011 at 8:12 AM

    To Antonio

    HEP bisector of CEB, AEC = DEB
    AEC + 1/2 CEB = DEB + 1/2 CEB

    ReplyDelete
  6. Antonio GutierrezJanuary 26, 2011 at 2:10 PM

    Problem #574. To c.t.e.o: Thanks for your comment. What is the reason for your statement HEP is the bisector of CEB (why are points H, E, and P collinear?). Thanks.

    ReplyDelete
  7. c .t . e. oJanuary 27, 2011 at 4:05 AM

    To Antonio,Thanks. H, E, P are never collinear
    HEP is the same bisector for vertcales angles
    We have to prove bisector of P // to bisector of CBE
    AC, DB meet at P1, bisector of CEB meet DP1 at P
    bisector of P1 meet CD at E1
    In ▲P1CE1 1/2 P1 + P1CE1 + CE1P1 = 180
    In E1EPP1 180- CE1P1 + E1EP + E1EP + PBE + 1/2P1=360
    from both => 2CE1P1 = 2E1EP => CE1P1 = E1EP
    as corresponding angle => P1E1//PE

    ReplyDelete
  8. Antonio GutierrezJanuary 27, 2011 at 4:23 AM

    Problem #574, To c.t.e.o: Thanks.

    ReplyDelete
  9. Sumith PeirisJuly 14, 2015 at 3:21 AM

    Let G cut BC at X and AD at Y. Then < A = < C. Hence Tr. FXY is isoceles and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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