Home
Class 11
MATHS
The median AD of the triangle ABC is bis...

The median AD of the triangle ABC is bisected at E. BE meets AC in F then AF : AC is

A

`3:4`

B

`1:3`

C

`1:2`

D

`1:4`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • PROPERTIES OF VECTORS

    AAKASH SERIES|Exercise PRACTICE EXERCISES|55 Videos

  • PROPERTIES OF VECTORS

    AAKASH SERIES|Exercise EXERCISE - I|31 Videos

  • PROPERTIES OF TRIANGLES

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL II) (Linked Comprehension Type Questions Passage -III:)|3 Videos

  • RATE MEASUREMENT

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE - 1) (LEVEL - 1) (STRAIGHT OBJECTIVE TYPE QUESTION)|43 Videos

Similar Questions

Explore conceptually related problems

The median AD to DeltaABC is bisected at E and BE is produced to meet the side AC in F. Show that bar(AF)=1/3bar(AC)

In a triangle ABC, AD is the median drawn on the side BC is produced to E such that AD = ED prove that ABEC is a parallelogram.

In an equilateral triangle ABC, AD bot BC meeting BC in D then AD^2 =…………..

In Delta ABC, if the median AD drawn through A is perpendicular to the side AC, then 3ca cos A cos C + 2a^(2) =

If D is the mid-point of the side BC of triangle ABC and AD is perpendicular to AC, then

ABC is a triangle and 'O', any point in the palne of the triangle. The lines AO, BO and CO meet the sides BC, CA and AB is D,E,F respectively. Then (OD)/(AD)+(OE)/(BE)+(OF)/(CF) equal to

O' is any point in the interior of a triangle ABC. OD bot BC , OE bot AC and OF bot AB , Show that AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2 .

AAKASH SERIES-PROPERTIES OF VECTORS-EXERCISE - II
  1. If the vector -bar(i)+bar(j)-bar(k) bisects the angles between the vec...

    Text Solution

    |

  2. If A(1, -1, -3), B(2, 1, -2), C(-5, 2, -6) are the vertices of a Delta...

    Text Solution

    |

  3. The median AD of the triangle ABC is bisected at E. BE meets AC in F t...

    Text Solution

    |

  4. If S is the circumcentre, G the centroid, O the orthocentre of Delta A...

    Text Solution

    |

  5. If S is the circumcentre, O is the orthocentre of Delta ABC then vec(O...

    Text Solution

    |

  6. The P.V.'s of the vertices of a DeltaABC are bar(i)+bar(j)+bar(k), 4ba...

    Text Solution

    |

  7. The P.V.'s of the vertices of a triangle are 2bar(i)+3bar(j)+4bar(k), ...

    Text Solution

    |

  8. DeltaABC be an equilateral triangle whose orthocentre is the origin 'O...

    Text Solution

    |

  9. The P.V.'s of A, B, C are bar(i)+bar(j)+bar(k), 4bar(i)+bar(j)+bar(k),...

    Text Solution

    |

  10. In a DeltaABC, P.V.'s of midpoints of AB, AC are bar(i)-bar(j)+bar(k),...

    Text Solution

    |

  11. The vectors of bar(AB)=3bar(i)+4bar(k) and bar(AC)=5bar(i)-2bar(j)+4ba...

    Text Solution

    |

  12. If D, E, F are the midpoints of BC, CA, AB of Delta ABC, then vec(A)D ...

    Text Solution

    |

  13. If D, E and F are respectively the mid-points of AB, AC and BC in Delt...

    Text Solution

    |

  14. If ABCD is a parallelogram then bar(AC)+bar(BD)=

    Text Solution

    |

  15. ABCD is a parallelogram, with AC, BD as diagonals, then overset(-)(AC)...

    Text Solution

    |

  16. OACB is a parallelogram with bar(OC)=bar(a), bar(AB)=bar(b)" then "bar...

    Text Solution

    |

  17. If the diagonals of a parallelogram are bar(i)+5bar(j)-2bar(k) and -2b...

    Text Solution

    |

  18. ABCD is a quadrilateral. E is the point of intersection of the line jo...

    Text Solution

    |

  19. ABCDE is a pentagon then bar(AB)+bar(AE)+bar(BC)+bar(DC)+bar(ED)+bar...

    Text Solution

    |

  20. ABCDEF is a regular hexagon. If bar(AB)+bar(AE)+bar(BC)+bar(DC)+bar(...

    Text Solution

    |