Find without sin,cos,tan.Parallelogram

$$ It is easy to notice that $\Delta\text{BCP}\sim\Delta\text{ABQ}$ , where $\angle BAD=\angle BCP$ , $\angle AQB=\angle BPC$ , and $\angle ABQ=\angle CBP$ . Therefore $$ $\begin{aligned} \frac{\overrightarrow{BQ}}{\overrightarrow{AB}}&=\frac{\overrightarrow{BP}}{\overrightarrow{BC}}\\ \frac{\overrightarrow{BQ}}{7}&=\frac{4}{5}\\ \overrightarrow{BQ}&=\frac{28}{5}\\ &=\boxed{5.6\text{ cm}} \end{aligned}$

As ABCD is a parallelogram then DC=AB=7.Now PC=DC-DP=7-4=3.Using Phythagorean theorem we get BC=5=AD(AS its a parallelogram).Now lets connect BD. Using Phythagorean theorem we get BD=4*sqrt 2.Now triangle BAD=14 (Heron's formula).In triangle BAD=1/2 * 5 * BQ=14 or BQ=(14 * 2)/5=5.6