Given two concurent lines BA, BD and a fixed point C, consider all triangles KBH, KH passing through C. Among all these triangles find the triangle having the minimal area.
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.
![[alogo]](alogo.png){kind=small}
A problem of Euclid ![[0_0]](EuclidMinimization_files/EuclidMinimization_0_0_0.png)
![[0_1]](EuclidMinimization_files/EuclidMinimization_0_0_1.png)
![[0_2]](EuclidMinimization_files/EuclidMinimization_0_0_2.png)