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Math Help - Triangle split in half angle-wise

  1. #1
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    Talking Triangle split in half angle-wise



    I'm searching for a simple expression for the distance CD in relation to BE (=2BD) in the picture above.

    1) D is midpoint on BE. BE and hence BD = DE are known.

    2) Angles BAC = CAE = BAE/2, that is, C represents the "angular midpoint" on BE as "seen from A". Those angles are known.

    3) F is the center of a circle. Points A, B and E lie on the perimeter of that circle so AF = BF is its radius.

    4) Angles BFD = BAE are known. Hence triangle BDF is completely known (because of right angle and BD are also known).

    Number 4 above is actually necessary because the circle describes all positions of point A which yield constant angle BAE. (See PlanetMath: circumferential angle is half the corresponding central angle) I would be very interested in knowing if there exists a similar simple relationship for all points A such that instead CD remains constant!

    The problem has implications for perspective. If you stand at A and look at BE, then BC will appear to be as long as CE. And I want to know what other points of observation A exist, which would remove the actual center D from the apparent center C with an equal distance CD. Alternatively, at least how CD relates to BD.

    I hope that you find the question a little interesting!
    Last edited by Optiminimal; November 3rd 2007 at 04:50 AM.
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  2. #2
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    Quote Originally Posted by Optiminimal View Post


    I'm searching for a simple expression for the distance CD in relation to BE (=2BD) in the picture above.
    1) D is midpoint on BE. BE and hence BD = DE are known.
    2) Angles BAC = CAE = BAE/2, that is, C represents the "angular midpoint" on BE as "seen from A". Those angles are known.
    3) F is the center of a circle. Points A, B and E lie on the perimeter of that circle so AF = BF is its radius.
    4) Angles BFD = BAE are known. Hence triangle BDF is completely known (because of right angle and BD are also known).
    Number 4 above is actually necessary because the circle describes all positions of point A which yield constant angle BAE. (See PlanetMath: circumferential angle is half the corresponding central angle) I would be very interested in knowing if there exists a similar simple relationship for all points A such that instead CD remains constant!

    ...
    Hello,

    I'm afraid but I believe there isn't a simple expression for the distance CD in relation to BE

    For a given circle, a given distance EB (and that means a constant angle \angle(EAB)) the angle bisector of \angle(EAB) passes allways through the point X. X is the intersection of EB and the perpendicular bisector of EB, that means point X is independent from the position of point A.
    Attached Thumbnails Attached Thumbnails Triangle split in half angle-wise-wklhalb_in_puz.gif  
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  3. #3
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    Ah, okey. Thanks for the clear answer!
    I will take a look at what solution I can find and see what I can make out of it.

    Btw, what software are you using to make images like that? I just use Sketchup which isn't really made for geometrical purposes.
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  4. #4
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    Quote Originally Posted by Optiminimal View Post
    ...
    Btw, what software are you using to make images like that?...
    Hi,

    have a look here: euklid dynageo homepage
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