Angle bisector theorem

Given : ABCD is a square , CF bisects $ \displaystyle \angle ACD$ , BPQ perpendicular CF .
Prove : DQ = 2PE .

$ \displaystyle \text{Proof}\ :\ \ \theta =\beta =\alpha =\gamma ={{22.5}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ In\ \Delta CQG\ \ and\ \ \Delta CPG\ ,$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta =\beta $ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \angle CGQ=\angle CGP={{90}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CG=CG\ \ (\ \text{common}\ \text{side}\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \Delta CQG\cong \Delta CPG\ (\ A.S.A\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ CQ=CP$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{CQ}}{{DQ}}=\frac{{BC}}{{BD}}\ \ (\ \because \ \text{Angle}\ \text{bisector}\ \text{theorem}\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{CQ}}{{DQ}}=\frac{1}{{\sqrt{2}}}\ .........(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{PE}}{{CP}}=\frac{{BE}}{{BC}}\ \ (\ \because \ \text{Angle}\ \text{bisector}\ \text{theorem}\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{PE}}{{CP}}=\frac{1}{{\sqrt{2}}}\ .........(2)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (1)\ \times \ (2)\Rightarrow \ \ \frac{{CQ}}{{DQ}}\ \times \ \frac{{PE}}{{CP}}=\frac{1}{2}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{PE}}{{DQ}}=\frac{1}{2}\ (\ \because \ CQ=CP\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \ DQ\ =2PE$