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Prove that if the bisector of any angle...

Prove that if the bisector of any angle of a triangle and the perpendicular bisector of its opposite side intersect, they will intersect on the circumcircle of the triangle.

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Let `O` be the circumcentre of the given `Delta ABC` .Then, the perpendicular bisector of `BC` passes throught the point `O`. Let it cut the circumcircle of `Delta ABC` at `D` and `BC` at `E`. Join `OB` and `OC`.
In order to show that the perpendicular bisector of `BC` and the bisector of `/_A` of `Delta ABC` intersect at `D`, it is sufficient to show that `AD` is the bisector of`/_A`.
Clearly, are `BC` makes `/_ BOC` at the centre and `/_A` at a point A on the remaining part of the circle.
`:. /_BOC = 2/_ a` ......(I)
In `Delta OEB` and `Delta OEC`, we have `/_ OEB= /_OEC = 90^(@)` .
Hyp. OB `=` hyp. OC `=` radius of the circle.
OE `=` OE (common)
`:. Delta OEB ~= Delta OEC`
`:. /_ BOE = /_ COE = /_ A ` `[ :' /_ BOC = 2 /_A]`.
Now, arc DC makes `/_A` at the centre and therefore it will make `(1)/(2) /_A` at the point `A` on remaining part of the circle.
`:. /_CAD = (1)/(2) /_A`
This show that `AD` is the bisector of `/_A`
Thus, the bisector of `/_ A` and the perpendicular bisector of side `BC`, intersect at a point `D` lying on the circumcircle of `Delta ABC`.
Hence, the result follows.
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