Fewpal
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In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC

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Let angle bisector of ∠A intersect circumcircle of ∆ABC at D. Join DC and DB. 

∠BCD = ∠BAD [Angles in the same segment] 

⇒ ∠BCD = ∠BAD 1/ 2 ∠A [AD is bisector of ∠A] ...(i) 

Similarly ∠DBC = ∠DAC 1/ 2 ∠A ... (ii) 

From (i) and (ii) ∠DBC = ∠BCD 

⇒ BD = DC [sides opposite to equal angles are equal] 

⇒ D lies on the perpendicular bisector of BC. 

Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ∆ABC Proved.

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