# Math Help - Another new riddle

1. ## 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!

2. 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.