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  1. #1
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    Geometry - Help

    In this figure Prove "Angle" RQS = 2 \times "Angle" BDC

    "Angle" QRD = "Angle" SRB,
    "Angle" RSD = "Angle" ASC and
    DRB, DSC, QSA are straight lines.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let \widehat{QRS}=\widehat{BRS}=x and \widehat{RSD}=\widehat{ASC}=y
    \widehat{QRS}=180-2x
    \widehat{RQS}=180-\widehat{QRS}-\widehat{QSR}=
    =180-2y-(180-2x)=2(x-y)

    \widehat{BDC}=\widehat{RDS}=180-\widehat{DSR}-\widehat{SRD}=
    =180-x-y-(180-2x)=x-y
    So, \widehat{QRS}=2\widehat{BDC}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by name101 View Post
    In this figure Prove "Angle" RQS = 2 \times "Angle" BDC

    "Angle" QRD = "Angle" SRB,
    "Angle" RSD = "Angle" ASC and
    DRB, DSC, QSA are straight lines.
    I'm like, totally in awe of red_dog right now.

    How can you do the problem if there is no point R in the figure??

    -Dan
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by name101 View Post
    DRB, DSC, QSA are straight lines.
    DRB is a straight line, so I assumed R is the point on DB.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by red_dog View Post
    DRB is a straight line, so I assumed R is the point on DB.
    Ahhhhhh! I hear the grasshopper now. (Bows in respect.)

    -Dan
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  6. #6
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    sorry i left that minor detail out. Quick to pick this one up.
    Very good work.
    Thanks again.
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