**Covid-19 has led the world to go through a phenomenal transition .**

**E-learning is the future today.**

**Stay Home , Stay Safe and keep learning!!!**

AB ≅ DE , BC ≅ EF and AC ≅ DF

∴ ΔABC ≅ Δ DEF by SSS

1) In triangle ABC, AD is median on BC and AB = AC.

Prove that ∠ABD = ∠ACD

Statements |
Reasons |

1) AB = AC | 1) Given |

2) AD is a median | 2) Given |

3) BD = DC | 3) By definition of median. |

4) AD = AD | 4) Reflexive (common side) |

5) ΔADC ≅ ΔADB | 5) By SSS postulate |

6) ∠ABD = ∠ACD | 6) CPCTC |

2) ΔABC and ΔDBC are two isosceles triangle on the same base BC and vertices of A and D are on the same side of BC. If AD is extended to intersect BC at P.

Prove that i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC.

⇒ AB = AC and BD = DC

Statements |
Reasons |

1) AB = AC | 1) Given |

2) BD = CD | 2) Given |

3) AD = AD | 3) Reflexive (common side) |

4) ΔABD ≅ ΔACD | 4) By SSS postulate |

5) ∠BAP = ∠CAP | 5) By CPCTC ∠BAD = ∠CAD |

6) AP = AP | 6) Reflexive(common side) |

7) ΔBAP ≅ ΔCAP | 7) By SAS postulate |

8) ∠APB = ∠APC | 8) CPCTC |

9) BP = CP | 9) CPCTC |

10) ∠APB + ∠APC = 180 | 10) These two angles are linear pair angles and they are supplementary |

11) 2∠APB = 180 | 11) Addition property |

12) ang;APB = 90^{0} |
12) Division property |

13) AP is the perpendicular bisector of BC | 13) By definition of perpendicular bisector and from (9) and (12) |

•

•

•

•

•

Home Page

**Covid-19 has affected physical interactions between people.**

**Don't let it affect your learning.**

GMAT

GRE

1st Grade

2nd Grade

3rd Grade

4th Grade

5th Grade

6th Grade

7th grade math

8th grade math

9th grade math

10th grade math

11th grade math

12th grade math

Precalculus

Worksheets

Chapter wise Test

MCQ's

Math Dictionary

Graph Dictionary

Multiplicative tables

Math Teasers

NTSE

Chinese Numbers

CBSE Sample Papers