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.