$$ABC$$ is a triangle, $$P$$ is a point on $$AB$$, and $$Q$$ is point on $$AC$$ such that $$\angle AQP = \angle ABC$$. Complete the relation $$${{area\,\,of\,\,\Delta APQ} \over {area\,\,of\,\,\Delta ABC}} = {{\left( {...} \right)} \over {A{C^2}}}$$$

In a $$\Delta ABC,\,\angle A = {90^ \circ }$$ and $$AD$$ is an altitude. Complete the relation $${{BD} \over {BA}} = {{AB} \over {\left( {....} \right)}}$$.