ABC is a triangle.sides ab and ac are produced.bisectors of exterior angle's meet at o.

Asked by armaanjain | 3rd Aug, 2008, 11:21: AM

Expert Answer:

Pl. check the question again.Is the triangle  ABC given iscoceles with AB=AC?

If it is, then following is the solution.

Let angle ABC=angle ACB=x (say)

Let D and E be the points on AB and AC produced such that CBD and BCE become the exterior angles.

So ,

 angle CBO=1/2*[180-x].. (because BO bisects the exterior angle CBD) .

angle BCO=1/2*[180-x]....( because  CO bisects the exterior angle BCE).

So BOC is also an isoceles triangle.

so we get BO=CO.

Now, consider triangles ABO and ACO,

AB=AC ..(given)

BO=CO ..( proved above)

AO=AO..( common side)

so we have triangle  ABO is congruent to triangle ACO.(SSS rule)

so,

 angle BAO= angle CAO (C.P.C.T)

Thus, we have proved that AO bisects angle BAC.

 

 

Answered by  | 3rd Feb, 2009, 05:50: PM

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