Inside the regular hexagon ABCDEF there is a point G so that <DGE=60, <CGD=90. DG intersects AB at X. Find AX/XB.
Proof:
G is intersection of the circles (FOE) and (DE) of centers K, L respectively. Let M, N be midpoints of EF, OM; NL is midline of the trapezoid DEMO, that is, NL=3R/4 (R = AB). Easily KN = OM/6=R.sqrt (3)/12, consequently NL/KN=3.sqrt (3), but triangles KNL and XCE are similar, having perpendicular sides, so CE/CX=3.sqrt (3), but CE=R.sqrt (3), thus CX=R/3, hence AX/XB=2, done.
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