Saturday, September 24, 2016

Geometry Problem 1264 Elements: Triangle, Exterior Angle Bisector, Circumcircle, Circle, Perpendicular, 90 Degrees, Concurrent Lines

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

Click the figure below to view more details of problem 1264.

Geometry Problem 1264: Triangle, Exterior Angle Bisector, Circumcircle, Circle, Perpendicular, 90 Degrees, Concurrent Lines.

Posted by Antonio Gutierrez at 11:33 AM
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5 comments:

  1. Peter TranSeptember 24, 2016 at 2:12 PM

    To Antonio
    Referring to problem 1264, The line DF cut arc AC at 2 points G1 and G2.
    I understand that the concurrency of BH, EG and AC will be true for both points G per the problem statement.
    Do I understand it correctly ? Please clarify.

    Peter Tran

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    Replies
    1. Antonio GutierrezSeptember 24, 2016 at 2:40 PM

      To Peter,
      Affirmative answer. The concurrency of BH, EG and AC will be true for both points G. Thanks.

      Delete
  2. Sumith PeirisSeptember 24, 2016 at 11:36 PM

    Let AC, EG cut at X
    < DBF = < EBA = < EGA = < ECA = @ say ....(1)

    Let < ECB = < EAB = < EGB = € ...(2)

    Now < ADB = < € since < BCA = @ + €
    So GXBD is concyclic
    Hence < XBG = < XDG = 90-@-€
    So < BHG = 90
    But < XHG = 90 so BXH are collinear points proving that AC, GE and and BH are concurrent

    Sumith Peiris
    Moratuwa
    Sri Lanka


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  3. APOSTOLIS MANOLOUDISSeptember 25, 2016 at 4:41 AM

    Problem 1264
    Suppose EG intersects AC at point P, it suffices to show that the BP is perpendicular to AG.
    Suppose BK is the internal bisector of <ABC (K=means the arc AC),then KB perpendicular in ED.So arcAE=arcEC .Is <FBD=(<CAB+<ACB)/2=90-<ABC, <GDB=90-<FBD=<ABC/2 and
    <CDB=<BCA-<CBD=(<ACB-<BAC)/2. But <PGB=<EGB=<ECB=<ACB-<ACE=<ACB-(<CEB+<CDE)=
    =(<ACB-<BAC)/2=<PDB. So the GPBD is cyclic.If BP intersect AG at H’ then <H’PG+<PGA=
    =<GDB+<EGA=<ABC/2+<ECA=<KBC+<EKC=90.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  4. Peter TranSeptember 29, 2016 at 5:45 PM

    https://goo.gl/photos/NQrpJBvDM9aABBvB8
    Sumith and Apostolis already showed concurrency of BH, EG and AC. I will show that point G1 will have the same property as point G.
    Let EG1 meet AC at P and BP meet AG1 at H1
    We will show that BP⊥AG1 ( see sketch)
    Let u= ∠ (EBA)= ∠ (ECA)= ∠ (EG1A)= ∠ (DBC)
    And v=∠ (ECB)= ∠ (EG1B)
    In triangle BCD since ∠ (ACB)= u+ v and ∠ (CBD)= u=> ∠ (BDC)= v
    In triangle CFD we have ∠ (FDC)= 90-u-v
    Since ∠ (PG1B)= ∠ (BDP)=v => BDG1P is cyclic and ∠ (PBG1)= ∠ (PDG1)= 90-u-v
    In triangle BH1G1 ∠ (PBG1)+ ∠ (AG1B)= 90-u-v+u+v= 90
    So BP⊥AG1

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