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Imagine a triangle with angles 132,36,12
I draw the exterior angles bisectors (132 and 12)
I want to prove that these bisectors are equal... Please take a look to the figure I want to prove that AM'=CM
I tried to use Stewart's theorem but I couldn't prove it..

Or let me ask my question Another way:
If we have a triangle that two exterior angle bisectors of that are equal, Can we say that our triangle is isosceles?
If we can not, please give me an example...

Alireza H
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  • if $AO = CO,$ then $\triangle AOC$ isoceles, in which case $\angle OAC = \angle OCA$ which it doesn't, so its not. – Doug M Jul 18 '17 at 18:45
  • @DougM Thanks,I edit my question. but let me ask my question another way: " If we have a triangle that two exterior angle bisectors of that are equal, Can we say that our triangle is isosceles ? " – Alireza H Jul 18 '17 at 18:50
  • Now you have a problem with your angles. If $AM'$ bisects $\angle MAC$ then $\angle M'AC = 24$ as on your first figure. But more importantly, you have constructed a triangle with more than $180^\circ$ in the two identified angles. (correcting for $\angle M'AC = 24$) – Doug M Jul 18 '17 at 19:00
  • Anybody can help!? – Alireza H Jul 19 '17 at 21:10

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