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Math Help - Another new riddle

  1. #1
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    Another new riddle

    This one is a lot easier though, but still fun to do! It took me more time to create it than to solve it.

    The question is simple:

    Prove that the ratio between the area of \triangle ACD and the area of \triangle ACB is equal to \frac{sin \angle ACD}{sin \angle ACB}.

    Have fun!
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  2. #2
    Junior Member
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    Montréal
    Posts
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    Answer:
    Spoiler:
    We create point F that goes from C to a point on DB. CF becomes the height of both triangles (ACD & ACB), which we will use to find the area.

    sinCDB = CF/CD
    CDsinCDB = CF (or h)

    Area of ACB = (AB(CDsinCDB))/2
    Area of ACD = (AD(CDsinCDB))/2

    Ratio of the areas: ((AB(CDsinCDB))/2) / ((AD(CDsinCDB))/2)
    Ratio of the areas: AB/AD

    Now with the rules of sines:
    AB: sinACB/AB = sinCAB/CD
    AB: (sinACB)/(sinCAB/CD)

    AD: sinACD/AD = sinCAD/DC
    AD: (sinACD)/(sinCAD/DC)

    Now we set up what we already have:
    AB/AD = (sinACB)/(sinCAB/CD) / (sinACD)/(sinCAD/DC)
    We know that sinCAB and sinCAD are the same, so we can replace them by CAB.
    AB/AD = (sinACB)/(sinCAB/CD) / (sinACD)/(sinCAB/DC)
    We know that CD and DC is the same line, so we can replace them by CD.
    AB/AD = (sinACB)/(sinCAB/CD) / (sinACD)/(sinCAB/CD)
    Since we just proved that the denominators are the same, we can simplify it to:
    AB/AD = sinACB/sinACD
    Voilà! Sorry for not using LaTeX, I was too lazy.
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