Saturday, May 9, 2020

Dynamic Geometry 1479: Triangle, Circumcircle, Angle Bisector, Perpendicular Bisector, Chord, Concyclic Points, Parallel Lines, Step-by-step Illustration

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Dynamic Geometry 1479: Triangle, Circumcircle, Angle Bisector, Perpendicular Bisector, Chord, Concyclic Points, Parallel Lines, Step-by-step Illustration, GeoGebra, iPad.

3 comments:

  1. Solution sent by c.t.e.o :

    1) Triangles EBM, BFL isosceles => B is center of the circle passes through M, E, F, L
    2) Ang EFM = Ang HKM => EF//HK
    Sketch for P1479
    https://photos.app.goo.gl/cDnnRqmh2su1pcW99

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  2. https://photos.app.goo.gl/KQZZUh3UnhQKozyPA

    Let u= ∠ (HAB)= ∠ (HEA)
    And v=∠ (KCB)= ∠ (CFK)
    And w=∠ (HKM)= ∠ (HLM)
    1. quad. AHBL is cyclic => ∠ (HAB)= ∠ (BLH)= u= ∠(BEL)
     EBL is isosceles and BE=BL
    Similarly we also have BM=BF=BE
    So M, E, F, G are concyclic with B as the center
    2. In quad. MEFL we have ∠ (EFM)= ∠ (MLE)= w
    In quad. HKLM we have ∠ (HKM)= ∠ (HLM)
    So ∠ (EFM)= ∠ (HKM) => EF//HK

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  3. < BLH = < BAH = < HEA = < BEL, so BE = BL
    Similarly BM = BF = BE = BL hence MEFL is concyclic with B as centre

    Therefore < MKH = < MLH = < MFE so EF // HK

    Sumith Peiris
    Moratuwa
    Sri Lanka

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