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  • 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 and DBC are triangles on the 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 bisector of BC.

Hence proved.

Posted by

infoexpert24

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