Challenging Geometry Problem ABCD is a square. F is a point on BD. CF bisects angle ACD. Q lies on CD. BQ is perpendicular to CF. If AC and BD intersect at E and BQ intersects AC at P, then prove that DQ = 2(PE).
I tried to use Midpoint theorem, but anyhow obstacles come on my way. Can anyone help?
Please do provide an euclidean approach, thank you.
 A: I'm writing this answer referring to Rory's diagram.
First off apply Menelaus theorem in $\triangle CED$ taking $BQ$ as the transversal. This will give you $$\frac{PE}{DQ}=\frac 12 \cdot \frac {PC}{QC}$$
Now since $\triangle CRP \cong \triangle CRQ$ we get $PC=QC$ and the result immediately follows.
A: Here is a coordinate approach. In this diagram I have taken the side of the square to have length one, and points $A,B,C,D$ have the obvious coordinates.

Using simple analytic geometry we get these facts:

*

*$\overleftrightarrow{AC}$ has the equation $y=x$.

*$\overleftrightarrow{BD}$ has the equation $y=1-x$.

*$E$ is the point $(\frac 12,\frac 12)$.

*$\measuredangle ACD=45°$.

*$\measuredangle DCF=22.5°$.

*$\overleftrightarrow{CF}$ has the equation $y=(\sqrt 2-1)x$.

*$\overleftrightarrow{BQ}$ has the equation $y=1-(\sqrt 2+1)x$.

*$Q$ is the point $(\sqrt 2-1,0)$.

*$P$ is the point $(1-\frac{\sqrt 2}2,1-\frac{\sqrt 2}2)$

*$DQ=2-\sqrt 2$.

*$PE=1-\frac{\sqrt 2}2$.

*$DQ=2\cdot PE$.

Let me know if you need the details on any of those points. Most are quite simple: the most difficult is finding $PE$.
