Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ABC is equal to BOC.


Given: ΔABC with AB = AC


In ΔABC,


AB = AC (given)


ACB = ABC (opposite angles to equal sides are equal)


1/2 ACB = 1/2 ABC (divide both sides by 2)


OCB = OBC …(1) (As OB and OC are bisector of B and C)


Now, in ΔBOC,


OBC + OCB + BOC = 180° (angle sum property)


OBC + OBC + BOC = 180° (from (1))


2 OBC + BOC = 180°


ABC + BOC = 180° (because OB is bisector of B)


180° - DBA + BOC = 180°


- DBA + BOC = 0


⇒∠DBA = BOC Hence proved.


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