Geometry Proofs ( Similarity of Triangles)
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In this section we will discuss Geometry proofs on similar triangles.Basic Proportionality Theorem( 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.
1) In triangle ABC, DE || BC and intersects AB in D and AC in E.
| Given : DE || BC To Prove thant : AD / DB = AE / EC Construction : Join BC,CD and draw EF ┴ BA and DG ┴CA. |
Proof :
| 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) |
2) In triangle ABC, DE || AC where D is a point on AB.F and E are any two points on BC, such that DF || AE.
| Given :DE || AC and DF || AE . Prove that : BF / FE = BE / EC |
 |
| 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) |
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Note : The converse of this theorem is also true.
If a line divides any two sides of triangle in the same ratio, then the line intersect must be parallel to the third side.
• Geometry proofs
• GeometryProof-1
• GeometryProof 2
• Proofs on Area of similar triangles
• Pythagorean theorem
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