# 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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