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Math Help - Help: two column proof involving triangles/properties of parallel lines!

  1. #1
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    Help: two column proof involving triangles/properties of parallel lines!

    I have no idea how to start this, let alone finish it. The worst part is I have to solve it by the end of tonight! Would one of you guys please help me?
    Oh yeah: can I assume triangle BST is an equilateral triangle?

    Solve using a two column proof:
    Given: Line segment AS is parallel to line segment BT;
    The measure of angle 4 is equal to the measure of angle 5
    Prove: Ray SA bisects angle BSR

    Help: two column proof involving triangles/properties of parallel lines!-proof.gif

    Thank you!
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  2. #2
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    By corresponding angles we know that 1 & 4 are congruent.
    By alternant interior angles we know that 2 & 5 are congruent.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by frostbite View Post
    I have no idea how to start this, let alone finish it. The worst part is I have to solve it by the end of tonight! Would one of you guys please help me?
    Oh yeah: can I assume triangle BST is an equilateral triangle?

    Solve using a two column proof:
    Given: Line segment AS is parallel to line segment BT;
    The measure of angle 4 is equal to the measure of angle 5
    Prove: Ray SA bisects angle BSR

    Click image for larger version. 

Name:	proof.gif 
Views:	246 
Size:	2.0 KB 
ID:	3549

    Thank you!
    No, BST is an isoceles triangle, but not necessarily equiliteral.

    Set angle 5 = x
    Angle 5 = Angle 4 (Given)
    Angle 4 = x

    Angle 3 = 180 - 2x (interior angles of a triangle)

    Angle 2 = Angle 5 = x (Parallel lines, alternant interior angles)

    Angle 1 = 180 - (Angle 2 + Angle 3)
    Angle 1 = 180 - (x + (180 - 2x))
    Angle 1 = x

    Since angle 1 = angle 2, the line bisects the angle
    Last edited by janvdl; July 8th 2007 at 02:50 PM.
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  4. #4
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    Hello, frostbite!

    Can I assume triangle BST is an equilateral triangle? . . . . NO!

    Solve using a two-column proof.

    Given: AS \parallel BT;\;\angle 4 \,=\,\angle 5

    Prove: SA bisects \angle BSR

    Click image for larger version. 

Name:	proof.gif 
Views:	246 
Size:	2.0 KB 
ID:	3549
    I'll let you supply the reasons.


    We see that: . \angle BSR \:=\:\angle1 + \angle2 .[1]

    \angle BSR is an exterior angle of \Delta BST.
    . . Hence: . \angle BSR \:=\:\angle 4 + \angle 5
    Since \angle4 = \angle5, we have: . \angle BSR \:=\:2\cdot\angle4 .[2]

    Equate [1] and [2]: . \angle 1 + \angle 2 \;=\;2\cdot\angle 4 .[3]

    Since \angle1 = \angle4 . (alt-int. angles),
    . . [3] becomes: . \angle1 + \angle 2 \;=\;2\cdot\angle1

    and we have: . \angle 2 \:=\:\angle 1


    Therefore, SA bisects \angle BSR.

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  5. #5
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    Wow! I didn't expect so many replies so fast...
    Thank you for the help!
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  6. #6
    Bar0n janvdl's Avatar
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    You can expect 5 replies a minute at this forum.
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  7. #7
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    That's exactly why this is the only forum that I've ever posted on more than twice. All thanks to a few dozen key people. Good job!
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