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Thread: Geometry/Angles

  1. #1
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    Geometry/Angles

    plz solve my question it is given that angle B= 2 times angle C and also AB=DC we have to prove that angle BAC=72° and AD is the bisector of angle A.



    tri.jpg

  2. #2
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    Quote Originally Posted by ramjmi View Post
    plz solve my question it is given that angle B= 2 times angle C and also AB=DC we have to prove that angle BAC=72° and AD is the bisector of angle A.



    tri.jpg
    To start:

    mBAC = 180° - 3*(mBCA)

    Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
    Last edited by mmm4444bot; 08-11-2012 at 02:36 PM. Reason: A matter of degrees :>
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Quote Originally Posted by ramjmi View Post

    plz solve my question

    We are here to help you solve your exercise. Please show us your effort (i.e., show some work, explain some reasoning, or ask a specific question).

    Here's a link to our FORUM GUIDELINES.

    Cheers

  4. #4
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    Quote Originally Posted by Subhotosh Khan View Post
    To start:

    mBAC = 180° - 3*(mBCA)

    Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
    Let angle BAC = x.
    Than angle ABC = 2x.
    and angle BAC = 180 - 3x.

    But, Now how to find the value x ??

  5. #5
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    Now let us look at those triangles.

    How does the vertex-angle bisector intersect the base - meaning if the line AD is the bisector intersecting base BC - what would be the ratio of BD and DC? If the ratio turns out to be that - does it prove that line is vertex-bisector?

    Have you thought through this line?
    Last edited by Subhotosh Khan; 08-12-2012 at 08:36 AM. Reason: removed unintentional smiley
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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