AoPS - Problem

let $AQ\cap BR=Y$ so $\angle AYB=180^\circ- \angle BAY- \angle ABY=180^\circ- \frac{2 \angle ABC}{3}- \frac{2 \angle BAC}{3}= 120^\circ$ and because of $AT,BT$ angle bisectors in $AYB$ we have $YT$ angle bisector. So we have $\angle BYT= \angle TYA=60^\circ$. Then \angle $\angle BYT= \angle TYA=\angle BYT= =60^\circ$. Because of $\angle TAQ= \angle QAR$ and $\angle BYT= \angle AYR$ triangles $ATY,ARY$ symetric triangles. So $\angle YTR= \angle YRT$. Because of $\angle BYT= 60^\circ$ we have $\angle YTR= \angle TRY=30^\circ$. Then $\angle RTB= 180^\circ- \angle TBR- \angle TRB= 180^\circ- \frac{ \angle ABC}{3}- 30^\circ= 120^\circ+ \frac{ \angle BAC}{3}$. Because of $0^\circ<\angle BAC<90^\circ$ we have $120^\circ<\angle RTB<150^\circ$. So we get result.....