Friday, December 13, 2013

Geometry Problem 940: Circle, Tangent, Secant, Chord, Parallel, Perpendicular, 90 Degrees, Midpoint

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 940.

Online Geometry Problem 940: Circle, Tangent, Secant, Chord, Parallel, Perpendicular, 90 Degrees, Midpoint

Posted by Antonio Gutierrez at 5:58 PM
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4 comments:

  1. nakaDecember 13, 2013 at 9:27 PM

    Angle ABC =angle BFC =angle AHC
    So ABHC concylic and since ABCO concylic by tangent property , we have ABHCO concylic
    Angle OHC = angle OCA = 90
    The second part is by the property of chord, straight forward
    QED

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  2. Anti_MainstreamDecember 13, 2013 at 10:45 PM

    Notice that BDCE is a harmonic quadrilateral, so the cross ratio of pencil (FE, FD ; FC, FB) = -1. So, if we intersect this pencil with line EA, we will obtain a set of harmonic points, but FB cuts EA at the point of infinity, thus (E, D ; H, P∞) = -1, and it follows that H is the midpoint of DE and it follows that OH ⊥ DE.
    Proved.

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  3. Peter TranDecember 13, 2013 at 11:39 PM

    http://img21.imageshack.us/img21/2474/uim3.png

    Since FB//ED => Arc(EF)=Arc(BD)
    We have ∠ (CHD)= ½(Arc(EF)+ Arc(DC))
    = ½(Arc(BD)+ Arc(DC))= 1/2Arc(BC)= ∠ (COA)
    So quadrilateral OCAH is cyclic => OH ⊥ HA and H is the midpoint of ED

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  4. Ivan BazarovDecember 15, 2013 at 8:39 PM

    Call <A=a. <DBA+<BAD=<EDF+<FDB, but <EDF=<DBA so <BAD=<FDB=b. <OCH=90-(b+90-a/2)=a/2-b=<OAH which makes HACO cyclic.

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