We consider a triangle $ABC$ whose angles are less then $120°$ and construct the equilateral triangles $ABC'$, $BCA'$ and $ACB'$, exterior to $ABC$. $I$ denotes the intersection of $(AA')$ and $(CC')$.
1) Show that $AA'=BB'=CC'$.
2) Show that $\widehat{BIC}=\widehat{BIA}=120°$.
3) Show that the lines $(AA')$, $(BB')$ and $(CC')$ intersect.
I have no idea how to attack this exercise. For point 3), it is clear that $I$ must be the point of intersection, but I do not know thow to prove any of these three steps.
Can anyone suggest how to proceed! Thank you in advance.
Could you please write about you prove (1) and (2) without using the conclusion of (3)?
Here's my solution:
First, just connect $CC'$ but not $AA'$ $BB'$, draw the circumcircle of $ABC'$, and let the circle intersect $CC'$ at $I$. It's easy to see that angle $BIC'=60^o$. So $A',B,C,I$ are on the same circle and $B',A,I,C$ are on the same circle. Because angle $AIC=120^o$, angle $A'IC=60^o, AIA'$ are collinear. So do $BIB'$.
So we prove (2) and (3) are right and as Jack D'Aurizio pointed out, you can read Fermat point - Wikipedia to prove (1) is right.