\xa0Basic Proportionality Theorem (Thales theorem):\xa0If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio.\xa0\tIn ∆ABC , if DE || BC and intersects AB in D and AC in E then AD AE —- = —— DB EC\tProof on 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 :\xa0In ∆ABC , DE || BC and intersects AB in D and AC in E.Prove that :\xa0AD / DB = AE / ECConstruction :\xa0Join BC,CD and draw EF ┴ BA and DG ┴ CA.\xa0\xa0\t

Statements Reasons 1) EF ┴ BA1) Construction2) EF is the height of ∆ADE and ∆DBE2) Definition of perpendicular3)Area(∆ADE) = (AD .EF)/23)Area = (Base .height)/24)Area(∆DBE) =(DB.EF)/24) Area = (Base .height)/25)(Area(∆ADE))/(Area(∆DBE)) = AD/DB5) Divide (4) by (5)6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC6) Same as above7) ∆DBE ~∆DEC7) Both the ∆s are on the same base and between the same || lines.\xa08) Area(∆DBE)=area(∆DEC)8) If the two triangles are similar their areas are equal9) AD/DB =AE/EC9) From (5) and (6) and (7)\t