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Math Help - Another Geometry Proof (Circles and Chords)

  1. #1
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    Another Geometry Proof (Circles and Chords)

    So the same situation as the last..but new proof:

    Given: Tangent line AE at B. Secant EC, which meets the circle at D.

    Prove: <ABC = <BDC and <EBD = <BCD

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  2. #2
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    Re: Another Geometry Proof (Circles and Chords)

    Quote Originally Posted by Volux View Post
    Given: Tangent line AE at B. Secant EC, which meets the circle at D.
    Prove: <ABC = <BDC and <EBD = <BCD
    I will give you three basic facts. You use them to do the rest.
    2m(\angle BED) = m(\text{arc}(CB)) - m(\text{arc}(BD))

    2m(\angle CBD) = m(\text{arc}(CD))

    2m(\angle EBD) = m(\text{arc}(BD))
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  3. #3
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    Re: Another Geometry Proof (Circles and Chords)

    By using the 2m(<BED) and etc..do you mean 2 of angle BED?

    And wouldn't m(arc(CB)) - m(arc(BD)) essentially mean just arc CD?
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    Re: Another Geometry Proof (Circles and Chords)

    Quote Originally Posted by Volux View Post
    By using the 2m(<BED) and etc..do you mean 2 of angle BED?
    And wouldn't m(arc(CB)) - m(arc(BD)) essentially mean just arc CD?
    The m stands for measure.

    Are you sure that you know enough geometry to tackle these problems?
    You seem to be short on the basics.
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  5. #5
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    Re: Another Geometry Proof (Circles and Chords)

    Yes I know what m means, though we aren't required to use them in class. We have never actually used 2 angles for one thing either.

    I know geometry somewhat, it just happens to be that I have an A- in the class and got the harder proofs in which I am having a hard time with.

    I'm not too great at where to start and justifications..or proofs in overall. Not great at them =/
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  6. #6
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    Re: Another Geometry Proof (Circles and Chords)

    I know arc BD = <BDC. and arc CD = <CBD.

    How am I to put those together though?
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  7. #7
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    Re: Another Geometry Proof (Circles and Chords)

    Would this work for the proof?:

    1. given stuff..
    2. <EBD = (1/2)(arc)BD 2. theorem used in the last proof I gave
    3. <BDC = (1/2)(arc)BC 3. I AM UNSURE..help??
    <DBC=(1/2)(arc)DC
    <BCD=(1/2)(arc)BD
    4. <ABC=(1/2)(arc)BC 4. theorem used in the last proof i gave
    5. <ABC=<BDC 5. substitution
    <EBD=<BCD


    Also can you help me out with this next proof? SAME image as before and same given, except:

    PRove: BE(SQUARED) = EC (TIMES) ED

    Hint: Use similar triangles, and the theorem from #2(the proof above)
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  8. #8
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    Re: Another Geometry Proof (Circles and Chords)

    Hi volux,
    Which triangles are similar.Define them and use proportions
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