Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB equal to AC and DB equal to DC. Show that AD is the perpendicular bisector of BC.

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC,  AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC. 

Answers (1)

 

Given : AD intersects BC at O

ABC, DBC are triangles on same base BC and

AB = AC, DB = DC

To prove : \angleAOB = \angleAOC = 90°

\angleBOD = \angleCOD = 90°

Proof : In \triangleABD and \triangleACD

AB = AC                     (given)

AD = AD                     (common)

BD = CD                     (given)

\therefore \triangleABD \cong \triangleACD      (By SSS congruence)

\angleBAO = \angleCAO         (by CPCT)

Now; In \triangleAOB and \triangleAOC

AB = AC                     (given)

AO = OA                     (common)

\angleBAO = \angleCAO         (from above)

\therefore \triangleAOB \cong\triangleAOC      (By SAS congruence)

\Rightarrow BO = OC

And, \angleAOB = \angleAOC

Then \angleAOB + \angleAOC = 180°

\angleAOB + \angleAOB = 180°                                  (from above)

\angleAOB = 90°, \angleAOC = 90°

and \angleAOB = \angleCOD, \angleAOC = \angleBOD          (Vertically opposite angles)

\therefore AO \perp r bisector of BC.

Hence proved.

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question