Saturday, May 24, 2014

Geometry Problem 1018: Right Triangle, Circles, Angle

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

Click the figure below to see the complete problem 1018.

Online Math: Geometry Problem 1018: Right Triangle, Circles, Angle

Posted by Antonio Gutierrez at 8:39 PM
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5 comments:

  1. UnknownMay 25, 2014 at 1:24 AM

    ∠DBE＝180。－∠BDE－∠BED
    ＝180。－1/2（180。－∠A）－1/2（180。－∠C）
    ＝180。－90。＋1/2∠A－90。＋1/2∠C
    ＝1/2（∠A＋∠C）
    ＝1/2（180。－∠ABC）
    ＝1/2（180。－90。）
    ＝45。

    ReplyDelete
  2. UnknownMay 25, 2014 at 3:35 AM

    Consider angle CBD as X ---> angle BAC = 2X (tangents rule)
    Consider angle ABE as Y ---> angle BCA = 2Y (tangents rule)
    From here we can see in triangle ABC that 2X + 2Y = 90 --> X+Y = 45
    In triangle BDE we get :
    angle BDE = 2Y + X
    angle BED = 2X + Y
    angle BDE + angle BED = 3Y + 3X = 135 THUS DBE = 45

    ReplyDelete
  3. Ivan BazarovMay 26, 2014 at 1:22 PM

    <ABE=<BCE/2, and <CBD=<BAD/2, so <EBD=90-<ABE-<CBD=90-(<BCE/2+<BAD/2)=90-90/2=45

    ReplyDelete
  4. Sumith PeirisJuly 12, 2015 at 10:25 AM

    3 Angles of Tr. BED =< ABC + 2.<EBD. Hence < EBD = 90/2= 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. MarcoApril 20, 2024 at 8:05 PM

    Let <ABD=<ABD=x
    <BAD=180-2x
    <BCA=180-(180-2x)-90=2x-90
    <CBE=[180-(2x-90)]/2=135-x
    <DBE=<ABD+<CBE-<ABC=x+(135-x)-90=45

    ReplyDelete

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