Tuesday, September 26, 2017

Geometry Problem 1347: Triangle, Angle Bisector, Congruence, Circumcircle, Tangent, Equilateral Triangle

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

Details: Click on the figure below.

Geometry Problem 1347: Triangle, Angle Bisector, Congruence, Circumcircle, Tangent, Equilateral Triangle.

10 comments:

  1. ∠BDC = ∠BCE = ∠BEC, so BDCE is concyclic.
    => ∠ADE = ∠ACB
    => ∠ADE = ∠EDC = ∠BDC = 60.

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    Replies
    1. Please, explain your first step. Thanks

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    2. I thought it is obvious so I skip the explanation.
      * ∠BDC = ∠BCE is because the triangle BDC & triangle BCA are similar under the power of circle at B (with C is the tangent point),
      ** ∠BCE = ∠BEC is because BE = CE so that triangle BCE is isos.

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  2. Let X be the foot of the perpendicular from E to AB and Y that of E to DC.

    Triangles EXD and EYD are congruent ASA.

    So EX = EY and hence triangles EXB and EYC are congruent right angle, hypotenuse, side.

    It follows that < DBE = < DCE and hence BCED is concyclic and so < C = < BCE = < EBC = < BDC = < BEC

    Therefore Tr. BCE is equilateral

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. To Sumith
      I have 2 comments :
      1. Your proof never use the given data " BC tangent to the circumcircle of triangle ADC at C "
      2. referring to line 4 of your solution "
      It follows that < DBE = < DCE and hence BCED is concyclic and so < C = < BCE = < EBC = < BDC = < BEC "
      Please explain how do you get < EBC=<BDC

      Peter Tran

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    2. Peter
      Your 2 comments are interconnected.

      Because of the tangency,
      <A = DAC = DCB and hence Tr.s ABC & CBD are similar and so
      < BDC = C.

      But < EBC = < ECB = C
      Hence < EBC = < BDC

      Hope this clarifies

      Sumith

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    3. Once you prove BCED is concyclic, without additional construction we can prove BEC as equilateral triangle by using alternate-segment theorem.

      Suppose say the m(BCA) = x => m(ADC) = 180-x => m(CDB) = x
      Since BCED is concyclic, m(CEB) = m(CDB) = x and hence BCE is equilateral triangle.

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    4. Repeating my response to Peter's comments

      1- I do, in fact,use the data on the tangency which gives < A = < BCD
      2 - As a result of the 1- above Tr.s of ABC and CBD are similar. Hence < BDC = <BCE = < EBC.

      Trust this clarifies

      Sumith

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    5. Thanks for the clarification

      Peter Tran

      Delete
  3. https://photos.app.goo.gl/UZHIJKvyxwGfqzUG3

    Let ∠ (ADE)= ∠ (EDC)= u
    Let X and Y are the projection of E over AD and DC
    Triangle BEX congruent to CEY => ∠ (XBE)= ∠ (ECY) => quadrilateral BDEC is cyclic
    Let DE cut circumcircle of triangle ADC at F .
    F will be the midpoint of arc AFC and ∠ (FAC)= ∠ (FCA)= u
    Since BC tangent to circle ADC => ∠ (AFC)= ∠ (ACB)= u
    So AFC is a equilateral triangle and u= 60 degrees
    So EBC is also a equilateral triangle

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