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If a line drawn through two sides of a triangle parallel to a third side, it divides those sides proportionally.
If a line drawn through two sides of a triangle parallel to a third side, it divides those sides proportionally.
In ∇ABC draw EF parallel to BC.
Let MB be contained m times in EB and n times in AE. Then
EB:AE = m:n
Draw lines parallel to BC and EF from the divisions on AB marked off by MB. These will create m intersections with FC and n with AF.
A line can be drawn through A parallel to EF.
These intersections with AC will also be of equal lengths to each other, although not necessarily the same length as the intersects on AB, as they are all equal intercepts of transversals between parallel lines.
∴ AF:FC = m:n = AE:EB
This explanation assumes that a value exists for MB that will divide into both EB and AE a whole number of times. This may not always be true. In this case consider making MB smaller and smaller, making m and n bigger numbers, until it the remaining part of the triangle left over approaches zero.
I am not too satisfied with this proof
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