Multiplication, the theory – by Thales’ theorem
The diagram can be simplified by using an acute triangle.
Proof of Thales theorem :
If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
Given : In ∆ABC , DE || BC and intersects AB in D and AC in E.
Prove that : AD / DB = AE / EC
Construction : Join BC,CD and draw EF ┴ BA and DG ┴ CA.
1) EF ┴ BA 1) Construction
2) EF is the height of ∆ADE and ∆DBE 2) Definition of perpendicular
3)Area(ADE) = (AD.EF)/2 3)Area = (Base .height)/2
4)Area(DBE) =(DB.EF)/2 4) Area = (Base .height)/2
5)(Area(ADE))/(Area(DBE)) = AD/DB 5) Divide (3) by (4)
6) (Area(ADE))/(Area(DEC)) = AE/EC 6) Divide (3) by Area(DEC)
7) ∆DBE ~∆DEC 7) Both the ∆s are on the same base and
between the same || lines.
8) Area(∆DBE)=area(∆DEC) 8) So the two triangles have equal areas
9) AD/DB =AE/EC 9) From (5) and (6) and (7)
Not only this but also AD/AB = DE/BC
I borrowed this from http://www.ask-math.com/basic-proportionality-theorem.html
Some adjustments, but the Thales theorem is well done. I liked it.