MathDB

4

Part of 2018 Dutch BxMO TST

Problems(1)

dutch angle chasing candidate, < AED + < ADO = 90^o wanted

Source: Dutch BxMO TST 2018 p4

8/2/2019
In a non-isosceles triangle △ABC\vartriangle ABC we have ∠BAC=60o\angle BAC = 60^o. Let DD be the intersection of the angular bisector of ∠BAC\angle BAC with side BC,OBC, O the centre of the circumcircle of △ABC\vartriangle ABC and EE the intersection of AOAO and BCBC. Prove that ∠AED+∠ADO=90o\angle AED + \angle ADO = 90^o.
geometryanglescircumcircleAngle Chasing