Thursday, May 2, 2013

Problem 873: Quadrilateral, Diagonal, Triangle, Angle Bisector, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 873.

Online Geometry Problem 873: Quadrilateral, Diagonal, Triangle, Angle Bisector, Congruence

Posted by Antonio Gutierrez at 9:52 AM
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8 comments:

  1. nakaMay 3, 2013 at 2:57 AM

    Reflect BD along AD to B'D
    Since AD is the angle bisector, A-C-B' collinear

    As DB = DC = DB', D is the center of circle(BCB')
    So angleBDB' = 2*angleBCA = 96.

    AngleBDA = AngleB'DA = (1/2)*AngleBDB' = (1/2)*96 = 48 = angleBCA
    Hence B,D,C,A concylic.

    AngleBAC + angleBDC = 180
    (5x + 5x) + (48 + x) = 180
    x = 12

    ReplyDelete
    Replies
    1. Peter TranMay 3, 2013 at 2:50 PM

      http://img13.imageshack.us/img13/2826/problem873.png

      Delete
  2. AnonymousMay 3, 2013 at 11:41 AM

    There is another solution, namely x=42/5, which occurs when the reflection of BD is CD.

    ReplyDelete
    Replies
    1. nakaMay 3, 2013 at 9:29 PM

      Could you please give your detail solution?

      Delete
  3. Peter TranMay 4, 2013 at 10:08 PM

    http://img23.imageshack.us/img23/8498/problem8731.png

    Let circumcircle of ABC cut AD at D’
    We have D’B=D’C ( chords of congruent angles) and DB=DC
    So DD’ is a perpendicular bisector of BC or D coincide to D’.
    To satisfy requirements of the problem, DD’ cannot be a perpendicular bisector of BC So D coincide to D’
    Quadrilateral ABDC is cyclic so 48+5x+5x+x=180 => x=12

    ReplyDelete
  4. AnonymousMay 19, 2013 at 6:52 PM

    Is there any other way?

    ReplyDelete
  5. Sumith PeirisAugust 29, 2015 at 8:22 AM

    Yes there is another way

    Draw perpendiculars DM & DN to AB & AC which can easily be proved to be equal. Hence Tr. BMD & Tr. CDN are congruent. So < MBD = < DCN = 6x which shows that ABCD is cyclic. Hence < BCD = 5x and adding the angles in Tr.,ACD which add upto 180 and solving we get x = 12

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. APOSTOLIS MANOLOUDISAugust 22, 2016 at 11:58 PM

    Problem 873
    The point D is the intersection of the perpendicular bisector BC and bisector <CAB. So by the theorem of South Pole the ABDC is cyclic, then <ABC=x,<DCB=<DAB=5x=<BCD.But
    <ABD+<ACD=180 or x+5x+48=5x=180 or x=12.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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