Extend BC to its double BE and project E parallel to the given lines to points F, I respectively. The minimal triangle is BFI. This because area(BIF) = area(BGJI) <= area(BKH), for every point H on the half-line BD. See the file HyperbolaConstruction.html for a discussion of the locus of point E', as the line KH varies passing through the fixed point C.

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