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

5 comments:

  1. ∠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。

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  2. 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

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

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  4. 3 Angles of Tr. BED =< ABC + 2.<EBD. Hence < EBD = 90/2= 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. 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