MODEL I 8. Thus. Given ARST with a point U inside such that <RST = LRTS-42°. and ZRSU - LSRU-18°. Find LUTS R PROBLEM I But Hence. Let Af be the midpoint of BC. AM intersects CP at Q. By the symmetry properties. we may have QBC (QCB = (PCB 3. LPBQ = LPBCQBC 35"- 30" - 5". and S M hence namely, BP bisects ADQ. Morover, since T 4PBQLABP. C (ADP ABC-PBC (180-(BAC)- <PBC 25. <AQP <<MQC = <MQB = 90° - 4QCM = 90 -<PCB Gr. <PQD = 2<BCQ = 2(DCP = 60°. <AQP = LPQB. namely, QP bisects <AQB. To sum up. Pis the incenter of AABQ. Thus. Ar bisects BAQ. It follows that <BAP: <BAM =<BAC 25".

Fig: 1