Geometry Proofs ( Similarity of Triangles)

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 :
Statements
Reasons
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
Proof :
Statements
Reasons
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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