Thursday, June 2, 2016

Geometry Problem 1221: Intersecting Circles, Chord, Tangent, Parallel Chords, Collinear Points

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 1221.

Geometry Problem 1221: Intersecting Circles, Chord, Tangent, Parallel Chords, Collinear Points

Posted by Antonio Gutierrez at 2:48 PM
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5 comments:

  1. Peter TranJune 2, 2016 at 6:17 PM

    Since EF//CD => Arc(EC)=Arc(FD)
    In circle O , We have ∠ (DTB)= ½( Arc(EC)+Arc(DB))= ½(Arc(FD)+Arc(DB))= ½ Arc(FB)= ∠ (BAF)… (1)
    Since DT tangent to circle Q at T => ∠ (BAT)= ∠ (DTB)…. (2)
    Compare ( 1) to (2) we have ∠ (BAF)= ∠ (BAT)
    So F, T , A are collinear

    ReplyDelete
  2. Sumith PeirisJune 2, 2016 at 11:23 PM

    < DTB = < FET but < DTB = < TAB

    So < FET = < TAB
    But < FET = < FAB
    Hence < TAB = < FAB

    So, F, T, A are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. APOSTOLIS MANOLOUDISJune 3, 2016 at 3:01 AM

    Is arc EC=arc FD. Suppose that AT intersects the circle with center O to the point Κ,then <EKA=<EBA=<ATC=<KTD so EK//CD.Therefore arc KD=arc FD, so the points K, F will coincide.
    Therefore points F,T,A are collinear.

    ReplyDelete
  4. AnonymousJune 3, 2016 at 9:13 AM

    Consider the line FA intersects the circle Q at P. Join AB.
    P and T coincide and hence F,T,A are collinear

    ReplyDelete
  5. c.t.e.o.October 6, 2020 at 12:39 PM

    Ang EFA= Ang EBA= Ang CTA, because EF//CD => F, T, A collinear

    ReplyDelete

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