Provide a proof for the following problem... How can I prove from this image that $BQ=2*PE.$ We know that $TM$ is parallel to $QD$ and that $CF$ is the bisector of angle $C$. As you see $QD$ and $TM$ are perpendicular to $CF$.
Obiously I found that $TQPE$ is an isoscel trapezoid, thus $TQ=EP$. The problem is I do not know how to prove that $BT=TQ$. 

Thanks in advance for your help! 
 A: As $TM$ is parallel to $QD$, then $DM=QT$. 
Examining the triangles $\triangle BTE$ and $\triangle EDM$, $\angle BET=\angle DEM$, $\angle TBE=\angle EDM$ and $\angle BTE=\angle EDM$, so the triangles are similar. 
In addition, $BE=ED$. So $BT=DM$, and as $DM=QT$, we have $BT=QT$.
A: Justify each step in the following:
It's all about angles, Thales Theorem (or similarity) and the angle bisector theorem. Denote
$$\angle BCF=\angle FCE=\alpha=22.5^\circ\;,\;\;\angle ACD=\beta=45^\circ$$
Look at the straight angle triangle $\;\Delta CFE\;$. Here, $\;\angle EFC=90^\circ-\alpha=:\gamma=67.5^\circ\;$ ,and from here:
$$\angle FET=90^\circ-\gamma=22.5^\circ=\alpha\implies \angle MED=\alpha\implies \angle PED=\alpha$$
But, as you say, it's easy to see that $\;TQPE\;$ is an isosceles trapezoid, and then
$$\angle TEC=\angle EC=\gamma\;,\;\;\angle TQP=\angle EPQ= 180^\circ-\gamma=112.5^\circ=\angle CPD$$
Finally (almost), look at the triangle $\;\Delta DEC\;$. Angles calculus here gives us $\;\angle PDC=\alpha\implies DQ\;$ is the angle bisector of $\;EDC\;$ here, and by the angle bisector theorem we get
$$\color{red}{\frac{CP}{PE}=\frac{CD}{DE}=\sqrt2}$$
But applying Thales theorem to $\;\angle TCE\;$ whose legs are intersected by the parallels $\;TE\,,\,\, QP\;$ , we get
$$\frac{CQ}{CT}=\frac{CP}{PE}=\sqrt2$$
and applying again the angle bisector theorem, but this time to $\;\Delta BCD\;$ and bisector $\;CQ\;$, we have
$$\frac{BQ}{QC}=\frac{BD}{CD}=\sqrt2$$
Complete now the exercise...
