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Math Help - Geometry puzzle

  1. #1
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    Geometry puzzle

    Code:
             A 
     
     
               E 
          D 
     
     
     
        C          B
    Triangle ABC is isosceles, equal angles at B and C = 80.

    D is on AC, such that CD = BC.

    E is on AB, such that angle BCE = 60.

    Find size of angle DEC. No trigonometry allowed! Have fun.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Wilmer View Post
    Code:
             A 
     
     
               E 
          D 
     
     
     
        C          B
    Triangle ABC is isosceles, equal angles at B and C = 80.

    D is on AC, such that CD = BC.

    E is on AB, such that angle BCE = 60.

    Find size of angle DEC. No trigonometry allowed! Have fun.
    DEC=30

    Full solution:
    Spoiler:
    I drew a scale diagram and measured it.
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  3. #3
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    BUT your "pencil" you "drew" with could be wide enough to be off a degree or so !!
    Nice try, Alex.
    But now that you have a diagram to play with, go "get the proof"
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  4. #4
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    1st hint: place point F on AB such that CF = CB
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  5. #5
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    2nd hint: examine triangle CFD
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  6. #6
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    Let  F be the reflection of  B across the foot of perpendicular of  C to AB . Then  CF = CB ,  \angle CFB = 80^o . We find that  \angle ECF = \angle CFB - \angle CEF = 80^o - 40^o = \angle CEF so  CF = EF  . At the same time ,  \angle DCF = 60^o and  CD = CB = CF so  \Delta DCF is equilateral . We thus have  FC = FD = FE . Therefore ,  F is the circumcentre of  \Delta DCE so  \angle CED = \frac{1}{2}~ \angle CFD = \frac{1}{2}~ 60^o = 30^o
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  7. #7
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    Nice. I like this solution (# means angle) :

    Triangle BCE: it is given that #ABC = 80 and #BCE = 60; then #BEC = 40

    Triangle CDF: this triangle being equilateral, then #CFD = 60;
    since #BFC = 80, this leaves #DFE = 180-80-60 = 40

    Triangle DEF: #EDF = #DEF, so #CED + 40 = 70, hence #CED = 30
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