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Homework Help: Geometrical problem

  1. Jul 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose ABC is a triangle, where ∠ABC = ∠ACB =80°. A line BD through B meets AC at D such that
    ∠DBC=60°.Similarly CE meets AB at E where ∠ECB = 50°. Whats the value of ∠EDB?

    2. Relevant equations

    3. The attempt at a solution
    I solved the problem using coordinate geometry but i was asked to solve it using simple geometry.
  2. jcsd
  3. Jul 11, 2013 #2


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    hi projjal! :wink:
    show us your diagram, with all angles marked, and all equal distances marked :smile:
  4. Jul 11, 2013 #3
    sorry for that. i have uploaded the diagram.

    Attached Files:

  5. Jul 11, 2013 #4
    Let's denote the intersection of EC and BD by F. Can you find the angles at this point F?

    Can you find the angle at A?
  6. Jul 11, 2013 #5


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    mark all the angles, and all equal distances

    (and preferably re-draw it so that those 80° angles look like 80°)
  7. Jul 11, 2013 #6
    I think now it should be okay with the diagrams.I have also marked all the angles.
    here also AB=AC & AD=BD
    now whats EDB ???

    Attached Files:

  8. Jul 11, 2013 #7


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    ah , that's much better! :smile:
    also BE = BC :wink:

    hmm … i can't see any obvious way to find EDB :redface:

    (longshot: i wonder whether it has anything to do with 40° being twice 20° ?)
  9. Jul 11, 2013 #8
    Draw a line in D parallel to BC.
  10. Jul 11, 2013 #9
    If you extend DE and CB until they meet at a point P, can you find angle EPB?
  11. Jul 11, 2013 #10
    I have already tried that but i couldnot find the answer.
  12. Jul 11, 2013 #11
    I couldnot get anything from that.If you could do it please specify it to me.
  13. Jul 11, 2013 #12
    EPDF form a quadrilateral, the other 3 angles are known.
  14. Jul 11, 2013 #13
    If P is the point of intersection of DE & CB how EPDF forms a quadrilateral?
  15. Jul 11, 2013 #14
    Sorry, EPBF - not enough screen area to see the diagram and type at the same time!
  16. Jul 12, 2013 #15


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    We know PBF and BFE. What's the third?
    I don't think it's going to be that simple anyway. Seems to me these angles have been chosen quite carefully. It's no coincidence that there are isosceles triangles all over the place which would not be so if we were to change just one of the given angles.
  17. Jul 12, 2013 #16
  18. Jul 12, 2013 #17
    So thats how its done.
  19. Jul 12, 2013 #18
    Oops, clearly not a good idea for me to try and do geometry in my head. I see the solution is not trivial.
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