# Geometry Proofs ( Similarity of Triangles)

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__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 |

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.

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**Geometry proofs**

• GeometryProof-1

• GeometryProof 2

• Proofs on Area of similar triangles

• Pythagorean theorem

• GeometryProof-1

• GeometryProof 2

• Proofs on Area of similar triangles

• Pythagorean theorem

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