Given arbelos A-C-B, D, E centers of (AC), (CB) respectively, while F, G are highest points of (AC), (CB) respectively. Prove that <AGD=<BFE.
Proof
Let V=AF x BG ( by homothecy it is highest point of (AB). Notice easy that FCGV is a rectangle, also if O is midpoint of AB, then DE=OA. Another interesting fact:
GV/FD=FC/FD=AV/AO=AV/DE=sqrt (2), meaning that triangles AGV and EFD are similar, so <VAG=<DEF, i.e. if T=AG x EF, <ATF=45 ( 1 )and T lies onto (AC). In a similar way, if U=BF x DG, U lies onto (CB) and <BUG=45 ( 2 ). From (1) and (2) we get FTUG cyclic, thus we are done.
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