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1) In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Prove that the line segments AF and EC trisect the diagonal BD.

Statements |
Reasons |

1) E and F are mid points of AB and CD respectively | 1) Given |

2) AE = 1/2 AB and CF = 1/2 CD | 2) Definition of mid point |

3) ABCD is a parallelogram | 3) Given |

4)AB =CD and AB || CD |
4) Properties of parallelogram |

5) AE = FC and AE || FC | 5) From (2) and (4) |

6) AECF is a parallelogram | 6) Properties of parallelogram and from (5) |

7) FA || CE and FQ || CP | 7) Properties of Parallelogram |

8) F is the mid point of CD and FQ ||CP | 8) By Mid Point -Theorem |

9) Q is the mid point of DP ⇒PQ = QD | 9) By Mid Point -Theorem |

10)Similarly, P is the mid point of BQ ⇒ BP = PQ | 10) By mid point theorem |

11) BP= PQ = QD | 11) From (9) and (10) |

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2) In ΔABC , right angled at B; and P is the mid point of AC. Prove that 1) PQ ⊥ AB 2) Q is the mid point of AB 3) PB = PA = ½ AC

P is the mid point of AC

Statements |
Reasons |

1) PQ || BC | 1) Given |

2) ∠AQP = ∠ABC | 2) Corresponding angles |

3) ∠AQP = 90^{0} |
3) Since ∠ABC =90^{0} |

4) ∠AQP + ∠BQP = 180^{0} |
4) Linear pair angles |

5) ∠AQP = ∠BQP = 90^{0} |
5) ∠AQP = 90 and from (4) |

6) PQ ⊥ AB | 6) From (5) |

7) P is the mid point of AC and PQ ||BC | 7) Given |

8) Q is the mid point. AQ =BQ | 8) By mid point theorem and definition of mid point. |

9) ∠AQP = ∠BQP | 9) From (5) |

10) PQ =PQ | 10) Reflexive (common) |

11) ΔAPQ = ΔBPQ | 11) SAS Postulate |

12) PA = PB | 12) CPCTC |

13) PA = 1/2 AC | 13) Since P is the mid point of AC |

• Introduction to Quadrilateral

• Types of Quadrilateral

• Properties of Quadrilateral

• Parallelogram and its Theorems

• Rectangle and its Theorems

• Square and its Theorems

• Rhombus and its Theorems

• Trapezoid (Trapezium)and its Theorems

• Kite and its Theorems

• Mid Point Theorem

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