Let $S=PQ\cap AT$, which must be the midpoint of $AT$. By PoP we have $SP\cdot SQ=SA^2=ST^2$, and so $P$ and $Q$ are inverses wrt $(AT)$. So, using parallel lines and the two similar triangles $STP$ and $SQT$, we have $\measuredangle BTQ=\measuredangle SQT=\measuredangle BTQ$.

Trivial. Let $M=AT\cap PQ$. Note that $MA^2=MT^2=MP\cdot MQ$, hence $\measuredangle PTA=\measuredangle PTM=\measuredangle MQT=\measuredangle BTQ$. [asy][asy] import olympiad;import geometry; size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,D,mb,mc,P,Q,M,T; O=(0,0);A=dir(120);B=dir(200);C=dir(340);path w=circumcircle(A,B,C);mc=midpoint(A--B);mb=midpoint(A--C); P=intersectionpoints(w,mb--100mc-99mb)[0];Q=intersectionpoints(w,mb--100mb-99mc)[0]; T=intersectionpoint(line(B,C),perpendicular(A,line(A,O))); M=extension(A,T,P,Q); draw(A--B--C--cycle,deep);draw(w,deep);draw(A--T--B,deep);draw(M--Q,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot(mb); dot(mc); dot("$M$",M,dir(M)); dot("$T$",T,dir(T)); [/asy][/asy]

Cheva sinus lemma and median lemma.

Nice problem

[i can't angle chase or pop help] Let $M$, $N$ be midpoints of $AB,AC$. Let $PQ\cap AT$ be $X$. By DIT on $AABC$ with line $MN$, we see that $\{M,N\}$, $\{P,Q\}$, $\{X,P_\infty\}$ are involutions ($MN$ is parallel to $BC$). Projecting this through $T$, $\{TM,TN\}$, $\{TA,TB\}$, and $\{TP,TQ\}$ are pairs of an involution. Now, clearly, $\triangle TAB\sim\triangle TCA$ and $M,N$ are corresponding midpoints so $TM$ and $TN$ are isogonal in $\angle BTA$. Clearly, $TA$ and $TB$ are also isogonal, hence this characterizes our involution, and $TP$, $TQ$ are isogonal as desired.