Trapezium Trigonometric Ratio

Let $h$ be the height of the trapezium. In $\Delta BAC$, height from C to AB is $h$, so $\sin\angle BAC = \frac{h}{AC}$. In $\Delta BAD$, height from D to AB is $h$, so $\sin\angle BAD = \frac{h}{AD}$. Ratio $\frac{\sin\angle BAC}{\sin\angle BAD} = \frac{h/AC}{h/AD} = \frac{AD}{AC}$.