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.