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

Posted by Antonio Gutierrez at 11:23 AM
Labels: , , , ,

11 comments:

  1. Peter TranJuly 13, 2014 at 4:27 PM

    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
    Replies
    1. AnonymousJuly 4, 2023 at 1:17 PM

      circumcenter of DBF will be on BC, (correct), E is center is not direct clear.

      Delete
    2. Peter TranJuly 9, 2023 at 10:41 AM

      circumcenter of DBF will be on BC, and circumcenter will be on perpendicular bisector of DF . the only satisfy these is point E

      Delete
  2. UnknownJuly 14, 2014 at 3:15 AM

    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. AnonymousJuly 15, 2014 at 8:42 PM

    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. AnonymousJuly 20, 2014 at 6:15 AM

    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. Antonio GutierrezJuly 20, 2014 at 9:41 AM

      To Bleaug
      Problem 1029 There is a condition
      CD greater than AD,
      Thanks

      Delete
    2. AnonymousJuly 20, 2014 at 2:17 PM

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

      Delete
    3. Antonio GutierrezJuly 20, 2014 at 2:46 PM

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

      Delete
  5. Sumith PeirisSeptember 2, 2015 at 3:55 AM

    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
  6. Sumith PeirisSeptember 2, 2015 at 4:06 AM

    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

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