Let ABCDEF be a regular hexagon 
Let $ABCDEF$ be a regular hexagon. Points $P$ and $Q$ on tangents to its circumcircle at $A$ and $D$ respectively such that $PQ$ touches the minor arc $EF$ of this circle. Prove that the angle between $PB$ and $QC$
is $30^o$. 
 A: 
(1) $ABP$~$DQC$ such that $\angle DCQ=\angle ABP$. Proof:
$$\angle CDQ=\angle BAP=150^o$$ due to contruction. Now, it sufficient to show that two triangles are similar if they have a common angle and the ratio of its sides are equal, i.e. $$\frac{AB}{DQ}=\frac{AP}{CD}$$
To show it we observe that $OAP$~$ODQ$ where $O$ is the circumcenter of $ABCDEF$.
Since $APTO$ and $ODQT$ are cyclic we have the following: $$\angle APO=\angle OAT$$ $$\angle QDO=\angle QTD$$ $$\angle OAT=\angle QTD$$ Because $QT=QD$, $OA=OT$ and since $\angle TOA=\angle TQB$ the above is shown.
So we have showed that $APO$~$QDO$ we have: $$\frac{AP}{OD}=\frac{AO}{DQ}$$ Now, $OD=CD$ and $AO=AB$ because it is well known that the sides of a regular hexagon is the same of the radius from the circumscribed circle.
Hence we have proved the proposition.
Now lets call $\angle APB=x$ and $\angle APT=y$. $$\angle CQD=30-x$$ $$\angle DQT=180-y$$
$$\angle GQP=\angle DQT-\angle DQC=180-y-(30-x)$$ $$\angle GPQ=\angle APT-\angle APB=y-x$$ $$\angle PGQ=180-(\angle GPQ+\angle GQP)=180-(y-x+180-y-(30-x))=30^o$$
And it's done!
