If AD AE
---- = ------ then DE || BC DB EC |

such that AD / DB = AE / EC.

Let DE is not parallel to BC. Then there must be another line that is parallel to BC.

Let DF || BC.

Statements |
Reasons |

1) DF || BC | 1) By assumption |

2) AD / DB = AF / FC | 2) By Basic Proportionality theorem |

3) AD / DB = AE /EC | 3) Given |

4) AF / FC = AE / EC | 4) By transitivity (from 2 and 3) |

5) (AF/FC) + 1 = (AE/EC) + 1 | 5) Adding 1 to both side |

6) (AF + FC )/FC = (AE + EC)/EC | 6) By simplifying |

7) AC /FC = AC / EC | 7) AC = AF + FC and AC = AE + EC |

8) FC = EC | 8) As the numerator are same so denominators are equal |

This is possible when F and E are same. So DF is the line DE itself.

∴ DF || BC

1) D and E are respectively the points on the sides AB and AC of a ΔABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, prove that DE || BC.

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

∴ BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And EC = AC – AE = 7.2 – 1.8 = 5.4 cm

Now, AD / DB =1.4 / 4.2 = 1/3

And AE / EC = 1.8 / 5.4 = 1/3

⇒ AD / DB = AE / EC

DE || BC ( by converse of basic proportionality theorem)

---------------------------------------------------------------

2) State whether PQ || EF.

DP / PE = 3.9 / 3 = 13/10

DQ / QF = 3.6 /2.4 = 3/2

So, DP / PE ≠ DQ / QF

⇒ PQ

• Similarity in Geometry

• Properties of similar triangles

• Basic Proportionality Theorem(Thales theorem)

• Converse of Basic Proportionality Theorem

• Interior Angle Bisector Theorem

• Exterior Angle Bisector Theorem

• Proofs on Basic Proportionality

• Criteria of Similarity of Triangles

• Geometric Mean of Similar Triangles

• Areas of Two Similar Triangles

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