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
Labels: , , , ,

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

Share your solution or comment below! Your input is valuable and may be shared with the community.

Subscribe to: Post Comments (Atom)