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Math Help - Circle Question

  1. #1
    s3a
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    Circle Question

    I don't know what to do on this question and would appreciate detailed help.
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    I don't know what to do on this question and would appreciate detailed help.
    Thanks in advance!
    triangles ADB and BDC are similar.

    so,

    \frac{AD}{BD}=\frac{DB}{DC}

    \frac{AD}{12}=\frac{12}{5}

    AD=\frac{144}{5}=28.8\;\;cm

    AC=AD+DC=28.8+5=33.8 \;\;cm

    ED = AC-(14+5)=33.8-19=14.8\;\;cm

    EC=AC-AE=33.8=14=19.8\;\;cm

    In triangle EBD,

    EB^2=BD^2+ED^2

    EB=\sqrt{(12)^2+(14.8)^2}

    EB = 19.05 cm

    NOW,

    FE \times EB=AE \times EC

    FE \times 19.05=14 \times 19.8

    FE = 14.6 cm

    got it ???
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  3. #3
    s3a
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    Thank you very much! I copied and analyzed your work for my assignment but even though the assignment is over with, I'd just like to ask one little question if you can still answer:

    How do you prove that triangles ADB and BDC are similar?
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