Given : DE || BC To Prove thant : 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 (4) by (5) |
6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC | 6) Same as above |
7) ∆DBE ~∆DEC | 7) Both the ∆s are on the same base and between the same || lines. |
8) Area(∆DBE)=area(∆DEC) | 8) If the two triangles are similar their areas are equal |
9) AD/DB =AE/EC | 9) From (5) and (6) and (7) |
Given :DE || AC and DF || AE . Prove that : BF / FE = BE / EC |
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1) DE || AC | 1) Given |
2) BD/AD = BE/AC | 2) Basic Proportionality Theorem |
3) DF || AE | 3)Given |
4) BD/AD = BF/FE | 4) Basic Proportionality Theorem in ∆ ABE |
5) BE/AC = BF/FE | 5) From (2) and (4) (Transitivity) |