Home
Class 9
MATHS
Given AD=DC and DB biscets angleADC. (...

Given `AD=DC` and DB biscets `angleADC`.
(i) Prove that, `DeltaADB cong DeltaCDB`
(ii) If `angleABD=48^(@)`, find `angleCBD`.

Text Solution

Verified by Experts

The correct Answer is:
`48^(@)`
Promotional Banner

Topper's Solved these Questions

  • TRIANGLES

    NAGEEN PRAKASHAN|Exercise Exercise 7b|15 Videos

  • TRIANGLES

    NAGEEN PRAKASHAN|Exercise Revision Exercise (very Short Answer Questions)|10 Videos

  • TRIANGLES

    NAGEEN PRAKASHAN|Exercise Problems From NCERT/exemplar|5 Videos

  • SURFACE AREA AND VOLUME

    NAGEEN PRAKASHAN|Exercise Revision Exercise (long Answer Questions)|10 Videos

Similar Questions

Explore conceptually related problems

In the adjoining figure, AD = DC and bisects angleADC . Prove that DeltaADB cong DeltaCDB .

Given a DeltaABD in which AB=AD and AC bisects BD . Prove that : DeltaABC cong DeltaADC .

In the given figure, ABCD is a quadrilateral in which AD = BC and angleDAB=angleCBA. Prove that (i) DeltaABD~=DeltaBAC, (ii) BD = AC, (iii) angleABD=angleBAC.

In a tringle ABC, AB = AC and bisector of angle A meets BC at D. Prove that : (i) DeltaABD cong DeltaACD (ii) AD is perpendicular to BC.

In the adjoining figure, BD = CE and angleADB=angleAEC=90^(@) , prove that (i) DeltaABD cong DeltaACE (ii) ABC is an isosceles triangle.

In the given figure, AB = CD and AD = CB. Prove that DeltaABD~=DeltaCDB .

In the given figure, OA = OB and OD = OC. Show that (i) DeltaAOD cong DeltaBOC and (ii) AD||BC

In the adjoining figure, angleBAC=angleBDC and angleABC=angleBCD . Prove that : (i) DeltaABC cong DeltaDCB (ii) DeltaABE cong DeltaDCF .

If ABCD is circular quadrilateral. If side AB and DC meets at point P on extension similarly AD and BC meet at point Q. If angleADC=85, angleBPC=40^(@) , find angleCQD .

In Fig. AC and AD are tangent to a circle at C and D respectively. If angleBCD = 44^(@) , then find angleCAD, angleADC, angleCBD and angleACD .

NAGEEN PRAKASHAN-TRIANGLES-Exercise 7a
  1. In the adjoining figure, angleBAC=angleBDC and angleABC=angleBCD. Prov...

    Text Solution

    |

  2. In the adjoining figure, ABCD is a quadrilateral. M and N are the poin...

    Text Solution

    |

  3. Given AD=DC and DB biscets angleADC. (i) Prove that, DeltaADB cong D...

    Text Solution

    |

  4. The adjoining figure shows a square ABCD and an equilateral triangle D...

    Text Solution

    |

  5. Equilateral triangles ABD and ACE are drawn on sides AB and AC respect...

    Text Solution

    |

  6. The following figure shows a square ABCD and an equilateral triangle D...

    Text Solution

    |

  7. The given figure showns a parallelogram ABCD. Squares ABPQ and ADRS ar...

    Text Solution

    |

  8. In a DeltaABC, BD is the median to the side Ac, BD is produced to E su...

    Text Solution

    |

  9. In the given figure, angleBDC=angleBEA and AB = BC. Show that AE=CD.

    Text Solution

    |

  10. If the diagonals of a quadrilateral bisect each other at right angle, ...

    Text Solution

    |

  11. In the adjoining figure, BM and DN are the perpendiculars from B and D...

    Text Solution

    |

  12. In the adjoining figure, AB=EF, BC=DE, angleABC=angleFED=90^(@), prove...

    Text Solution

    |

  13. In the adjoining figure, ABCD is a parallelogram. If angleMBC = angleN...

    Text Solution

    |

  14. In the adjoining figure, QX and RX are the bisectors of the angles Q a...

    Text Solution

    |

  15. In the following figure, OA = OC and AB = BC. Prove that : (i) angle...

    Text Solution

    |

  16. In the adjoining figure, ABCD is a paralogram. The side AB is produced...

    Text Solution

    |

  17. In the following figures, AB=PQ, AC=PR and AM=PN. Prove that DeltaABC ...

    Text Solution

    |

  18. Prove that the medians of an equilateral triangle are equal.

    Text Solution

    |

  19. ABCD is a square EF is parallel to BD. R is the mid-point of EF. Prove...

    Text Solution

    |

  20. The following figure shows a triangle ABC in which Ab = AC. M is a poi...

    Text Solution

    |