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

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

Answers (1)

Given: The angle bisector of any angle of a triangle and the perpendicular bisector of the opposite side intersect on the circumcircle of the triangle.

Consider DABC

Let the angle bisector of angle A intersect the circumcircle of triangle ABC at point D.

Join DC and DB

\angleBCD = \angleBAD  (Angle in the same segment are equal)

\angleBCD = \angleBAD = \frac{1}{2} \angleA      … (1)   (AD is bisector of ÐCAB)    

Similarly,

\angleDAC = \angleDBC  (Angle in the same segment are equal)

\angleDBC =\angleDAC = \frac{1}{2} \angleÐA       … (2)   (AD is bisector of ÐCAB)    

From (1) and (2) we have

\angleDBC =\angleBCD

BD = DC (sides opposite to equal angles in a triangle are equal)

Hence, D also lies on the perpendicular bisector of BC.

Hence, proved

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