Question

prove that the angle bisector and the perpendicular bisector of the opposite side if intersect they will intersect on the circumcircle of the triangle.

Answer 1

Let O is the circumcenter and ABC is the triangle.
Let the bisector of angle A and the perpendicular bisector of BC intersect at D.
Since angle subtended by the arc at the center is twice the angle subtended by the arc at the point of the alternate segment of the circle.
So, From the figure,
∠BOC = 2∠A
Again, OB = OC   {Radii of the circle}
So, ΔBOC is an isosceles triangle.
=> ∠BOC = ∠COE = ∠A  {since OD is perpendicular bisector of BC}
Since any three points are always concyclic
So, A, C and D are concyclic.
Now, arc CD subtends A/2 at A and ∠A at O.
So, O is the center of the circle passing through A, C and D.
Hence, the circle passing through A, C and D is the circumcircle of triangle ABC.
Similarly, the circle passing through A, B and D is the circumcircle of triangle ABC.
Thus, P lies on the circumcircle of triangle ABC.
Hence, 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

Answer 2

Step-by-step explanation:

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