Sunday, October 30, 2016

Geometry Problem 1281 Triangle, Circle, Diameter, Tangent, Altitude, Congruent Angles

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 1281.

Geometry Problem 1281: Triangle, Circle, Diameter, Tangent, Altitude, Congruent Angles.
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4 comments:

  1. Peter TranOctober 30, 2016 at 6:09 PM

    https://goo.gl/photos/cZVha9LddjDaWn519

    Connect OB, OD ,OE where O is the midpoint of AC
    We have OD⊥BD and OE⊥BE and OB is the angle bisector or angle BDE
    So O, H, D, B, E are cocyclic
    In circle OHDBE ∠AHD= ∠DBO
    ∠CHE=∠OHE= ∠OBE
    But ∠DBO=∠OBE => ∠AHD=∠CHE

    ReplyDelete
  2. Sumith PeirisOctober 30, 2016 at 9:55 PM

    Let AD, HB meet at P. Let HPB extended meet AD extended at X.

    Now P is the orthocenter of triangle AXC, hence AP is perpendicular to XC.
    But AE is perpendicular to XC. It follows that A,P,E are collinear and XEC
    are collinear.

    P being the orthocenter of triangle AXC, easily P is the incentre of
    triangle HDE.

    So < AHD = < CHE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. APOSTOLIS MANOLOUDISOctober 30, 2016 at 11:28 PM

    Problem 1281

    Suppose that the DC intersect BH in point P and AD the BH to Q.Then < DCA=<QDB ( cord and tangent) and <DQB=<DCA (perpendicular),so <DQB=<QDB or BD=BQ=BE. But <PDB=90-<BDQ=<DPB,then BD=BP=BE=BQ. So PE is perpendicular to BE.
    Then at triangle PAC the point P is orthocenter, and A,P and E
    Are collinear. The ADPH ,HPEC are concyclic. So <DHA=<DPA=<EPC=<EHC.
    4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE

    ReplyDelete
  4. AnonymousNovember 1, 2016 at 10:31 AM

    m(BOD) = mBOE) = m(BOH) = 90 => B,D,H,O & E are concylic
    Since BD = BE => m(BHD) = m(BHE) = @ (say)
    Therefore m(AHD) = m(CHE) = 90-@

    ReplyDelete

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