In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
In trapezium $ABCD$, side $AB \parallel PQ \parallel DC$. If $AP = 3$, $PD = 12$, and $QC = 14$, find $BQ$.