Sunday, July 13, 2014

Geometry Problem 1029: Right Triangle, 90 Degrees, Angle, Midpoint, 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 1029.

Online Math: Geometry Problem 1029: Right Triangle, 90 Degrees, Angle, Midpoint, Congruence

10 comments:

  1. Extend DE to F so that ED=EF => BDCF is a parallelogram
    Note that ∠(FDC)= ∠(DFB)= ∠(DBA) => BA is a tangent to circumcirle of triangle DBF
    Circumcenter of DBF will be on BC => E is the circumcenter of DBF
    So EB=ED=EC => BD perpendicular to AC

    ReplyDelete
  2. Is this solution is correct? :
    1)Continue DE till point F such that BE = CE = EF --> BFC right triangle and E is the circumscribed circle
    of BFC and since BC is the diameter And angle ABC is right so AB is tanget and so on ....

    ReplyDelete
  3. Hi Antonio

    Please publish full of my problema.

    Prob 1027
    We build AH perpendicular BC and DF perpendicular AB
    CH=HD=1/2DC=1/2ED
    DF=DH=1/2ED
    α =10
    X=90-4α
    X=50

    Prob 1028

    BE=ED=EC EM=EN EN =1/2 EC
    Triangle ECN < ECN =30 ECN = 90-4X -> X =15

    Prob 1029
    We note <BDF = 90
    BD perpendicular AC

    Erina New Jersey

    ReplyDelete
  4. as stated, problem 1029 is wrong.

    Given right ABC and BE=CE: ∠(DBA)=∠(EDC) does not imply BD⊥AC

    Take D' on AC such as ED'⊥BC, then ∠(D'BA)=∠(ED'C) but BD' is not ⊥AC

    bleaug

    ReplyDelete
    Replies
    1. To Bleaug
      Problem 1029 There is a condition
      CD greater than AD,
      Thanks

      Delete
    2. naahhh, seriously?
      then you should add this in the "Given:" section
      bleaug

      Delete
    3. To Bleaug
      Now it has been included in the "Given" section. Thanks

      Delete
  5. Thanks Antonio for the clarification.

    Above solutions do not make explicit use of this newly added premiss, therefore they must be incomplete at the very least. Without it, there are two solutions to the problem, only one obeying CD>AD. For example in Peter's solution, there is no justification of the implication on the 3rd line.

    Background: if we make D vary on AC and let F on ED be such that ∠(EDC) = x = ∠(FBA), it is easy to show that ∠(BFE) is constant and equal to ∠(BAC). See http://bleaug.free.fr/gogeometry/1029.png

    Therefore, the locus of all F when D varies lies on the circle circumscribed to FEB. This circle crosses AC in at most two points. With simple geometric arguments we can show that the two following points belong to both AC and circle FEB and thus can be candidate solutions: D1 such as BD1⊥AC and D2 midpoint of AC. Since AD>CD, only D1 is acceptable => BD⊥AC

    bleaug

    ReplyDelete
  6. Draw FE perpendicular to BC to meet BD extended at F.

    Since AB // FE < EDC = < ABF = < BFE = < CFE since FE is the perpendicular bisector of BC. So CEDF is cyclic and BD is thus perpendicular to AC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Another method - let GE be the perpendicular bisector G on AC. Show that < CAB = ABG = < BGE = < CGE. Hence < DEG = < DBG , hence BDGE is cyclic and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete