KOMATH

<Tri-2026-2309>, hint

<Tri-202-2309>,general, center

<Tri-2026-2308>, hint

Suppose AM cuts EF at K'

Since [ABM]=[ACM] ;ABxAM sinBAM=ACxAM sin CAM ; 2RsinC sinBAM=2RsinB sinCAM

sinBAM/sinCAM=sinB/sinC Hence FK'/EK'=sinB/sinC

2[DKF]=DFxDK sinB/2=2IF cosB/2 xDK sinB/2=IFx DK sinB

Similarly 2[DKE]=IEx DK sinC. IE=IF ( inradius), [DEK]/[DFK]=EK/FK

Hence EK/FK=sinC/sinB=EK'/FK'. Hence K=K' A,K,M are collinear.

<Tri-2026-2308>, general, collinear

<Cir-2026-800>, hint

Since CPQ is isosceles PQ ㅗCO (O;enter of circle ABC) Triangles CLA and CNA are congruent (KAxKC=KPxKQ=KLxKN, A,N,C,L are concyclic <CLA=<CNA=90,CL=CN, AC common)

<COB=2<CAB=<LAB hence AL//CO and COㅗPQㅗAL