June 25th, 2017, 01:54 AM  #1 
Newbie Joined: Jun 2017 From: Viet Nam Posts: 2 Thanks: 0  Geometry
In a triangle $ABC\!$, let $A_0$ be the point where the interior angle bisector of angle $A$ meets with side $BC\!$. Similarly define $B_0$ and $C_0$. Prove that $\angle B_0A_0C_0 = 90^\circ$ if and only if $\angle BAC = 120^\circ\!$. p/s: please help me. I have posted this in many forums, but get no answer. Source: 2017 Math Majors of America Tournament for High Schools Tiebreaker Round problem 4. Last edited by skipjack; June 25th, 2017 at 04:55 AM. 
June 25th, 2017, 06:44 AM  #3  
Newbie Joined: Jun 2017 From: Viet Nam Posts: 2 Thanks: 0  Quote:
That solution just proves 1 side, I mean how to prove the other side. Last edited by skipjack; June 25th, 2017 at 08:23 AM.  
June 26th, 2017, 02:36 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,421 Thanks: 1462 
Are you given that $AB = AC$? Did you try to prove the generalization given at the end of the linked page? 

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