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
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.
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,
BO=CO ..( proved above)
AO=AO..( common side)
so we have triangle ABO is congruent to triangle ACO.(SSS rule)
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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