Saturday, April 16, 2016

Geometry Problem 1209: Triangle, Circle, Excircle, Excenter, Circumcircle, Congruence

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 1209.

Geometry Problem 1209: Triangle, Circle, Excircle, Excenter, Circumcircle, Congruence

Posted by Antonio Gutierrez at 7:26 PM
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7 comments:

  1. Peter TranApril 16, 2016 at 8:43 PM

    Let F is the point of tangent of BC to circle E
    Since E is center of excircle of triangle ABC => ∠ (ACE)= ∠ (ECF)
    ∠ (ACE) subtend arc DE
    ∠ (ECF) subtend arc EC+ arc BC= arc BE
    Since ∠ (ACE)= ∠ (ECF)=> arc DE= arc BE=> DE=BE

    ReplyDelete
  2. APOSTOLIS MANOLOUDISApril 17, 2016 at 7:15 AM

    Is <CBE=<DCE=<ECF (EF⊥BC).But <ECF=<BDE.Therefore <DBC=<BDE or BE=DE.

    ReplyDelete
  3. c.t.e.oApril 17, 2016 at 8:06 AM

    Draw DM tg to E, N tg point of BC =>
    BE bisector, DE bisector =>
    R tr EDM congr ENB

    ReplyDelete
  4. Sumith PeirisApril 17, 2016 at 12:56 PM

    < BEC = 90 - C/2 - B/2 = A/2 = < BDC = < DBA since < BAC = A

    Further easily < EDC = < EBC = B/2 and so < BDE = A/2 + B/2 = < DBE and so BE = DE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Sumith PeirisApril 17, 2016 at 1:00 PM

    Further AD = AB as well

    ReplyDelete
  6. Sumith PeirisApril 17, 2016 at 1:03 PM

    Much simpler proof.....

    < DBE = < DCE = < BDE and the result follows

    ReplyDelete
  7. AnonymousMay 29, 2016 at 9:54 PM

    Let F be the point of tangent of BC to circle E on line BC
    ∠DBE=∠DCE=∠ECF=∠BDE
    ∴BE=DE

    ReplyDelete

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