Sunday, September 11, 2016

Geometry Problem 1259: Triangle, Circumcircle, Circle, Diameter, Parallel Lines

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 1259.

Geometry Problem 1259: Triangle, Circumcircle, Circle, Diameter, Parallel Lines.

Posted by Antonio Gutierrez at 2:36 PM
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3 comments:

  1. Sumith PeirisSeptember 11, 2016 at 3:46 PM

    BEFD is cyclic hence < CBD = < DEF
    ABCD is cyclic hence < CBD = < DAC

    So < DEF = < DAC

    Therefore EF //AC

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Peter TranSeptember 11, 2016 at 5:32 PM

    https://goo.gl/photos/3nftCTs6WqmHt8pAA
    Connect BD, BE and DF
    Note that BE⊥AD and DF⊥BC
    So Quadrilateral BEFD is cyclic => ∠ (BDE)= ∠ (BFE)= u
    Qua. BACD is cyclic => ∠ (BDA)= ∠ (BCA)= u
    So ∠ (BCA)= ∠ (BFE)= u => EF//AC

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  3. APOSTOLIS MANOLOUDISSeptember 12, 2016 at 1:00 AM

    Problem 1259
    Let K center of the circle with diameter AB and L center of the circle with diameter CD.
    Let <BAD=x=<BCD and <ABE=90-x=<FDC .Then <EBF=<ABC-<ABE=<ADC-<FDC=<EDF.
    So the EBDF is cyclic, then <FED=<FBD=<CBD=<CAD.Therefore EF//AC.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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