Wednesday, November 16, 2011

Problem 688: Triangle, Angles, 10, 20, 30, 40, 60 Degrees, Measure, Mind Map, Polya

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 688.

Online Geometry Problem 688: Triangle, Angles, 10, 20, 30, 40, 60 Degrees, Measure, Mind Map, Polya.

Posted by Antonio Gutierrez at 5:02 PM
Labels: , , , , , , ,

12 comments:

  1. AlejandroNovember 16, 2011 at 10:26 PM

    http://img403.imageshack.us/img403/6706/sfvdf.png

    Extend AB from B till F so that < EFD = 30°. Since < CEF = 50°, then < CDF = 50° and DF ⊥ BC, whereat < BDF = 30°, and finally from the cyclic quadrilateral CDEF we have x + 30° = 50° and x = 20°.

    ReplyDelete
  2. AnonymousNovember 18, 2011 at 3:43 PM

    Extend AB till F so that AF=AC. This way DCFE is an isosceles trapezoid(CD=CF=EF and <DCF=<CFE=80). <AED =80=<EBD+<BDE. So <BDE=20.

    ReplyDelete
  3. SCPDecember 22, 2011 at 2:03 AM

    Did you both used AD=AE which isn't given or how can someone conclude EF=CD?

    ReplyDelete
  4. Jacob HA (EMK2000)October 19, 2012 at 8:20 AM

    Reflection in the line BC. Let D→D'.

    ∵ ∠DBC = ∠D'BC = 60°
    ∴ E, B, D' are collinear.

    ∵ ∠BDC = ∠BD'C = 80°
    ∴ ∠D'CA = 80°, ∠D'CB = 40°

    Join DD'. Then BD = BD',
    ∴ ∠BDD' = 30°, ∠D'DC = 50°

    ∵ ∠D'DC = ∠D'EC = 50°
    ∴ C, D, E, D' are concyclic.

    ∴ ∠D'DE = ∠D'CE = 50°
    ∴ x = ∠BDE = ∠D'DE − ∠BDD' = 20°

    ReplyDelete
  5. Jacob HA (EMK2000)October 19, 2012 at 4:57 PM

    Reflection in the line CE. Let D→D'.

    Then ΔDCD' is equilateral.

    ∵ ∠DBC = ∠DD'C = 60°
    ∴ B, D, C, D' are concyclic.

    ∵ ∠EBD = ∠DCD' = 60°
    ∴ E, B, D' are collinear.

    In ΔACD', ∠AD'C = 100°
    ∴ ∠ED'D = 40°

    ∵ ED = ED'
    ∴ ∠EDD' = 40°

    ∵ ∠BDD' = ∠BDD' = 20°
    ∴ x = ∠EBD = 20°

    ReplyDelete
  6. Sumith PeirisAugust 25, 2015 at 1:48 AM

    Find F on AB extended such that BC bisects < ACF. Then Tr. EFC is congruent with Tr. DFC and so EC = EF from which we can prove that Tr.s AFD and ACE are congruent. Hence AE = AD and so x = 20

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisNovember 10, 2015 at 10:21 AM

    Or simply EF = FC = DC and CDEF is concyclic so x+30 = 10+40 hence x= 20

    ReplyDelete
  8. APOSTOLIS MANOLOUDISJune 27, 2016 at 3:53 AM

    Problem 688
    Draw the equilateral triangles DEK, then <EBD=<EKD=60.So EBKD is cyclic.Then <DEK=<DBK=60=<DBC.Therefore B,K,C are collinear. Is <EKD=60=2.30=2.<ECD,
    Then the point K is circumcenter of triangle DEC .Then x=<EDB=<EKB=10+10=20.
    MANOLOUDIS APOSTOLIS FROM GREECE

    ReplyDelete
  9. Geek37November 6, 2017 at 11:02 AM

    Here's my solution

    https://www.youtube.com/watch?v=6DCTArdbSpE

    ReplyDelete
    Replies
    1. Sumith PeirisNovember 18, 2017 at 6:09 AM

      You draw a parallel to AC thro' B and then u draw a circle (D,DB) to intersect the parallel at say P.

      Your proof then assumes that D,E,P are collinear -this is by no means evident or axiomatic. It must be PROVED.

      Antonio -I would like u to pls comment too

      Rgds

      Sumith Peiris

      Delete
    2. Geek37February 7, 2018 at 12:25 PM

      When I constructed the original drawing on Geogebra without any extra parts, the measurement of the angle was 20 degrees.

      Delete
    3. Sumith PeirisSeptember 29, 2018 at 7:02 AM

      We are asked to PROVE it using Euclidean principles

      Delete

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