# Show that $AB=AA_1$

Let $$ABC$$ a triangle with $$A_1$$ the middle of the edge $$[BC]$$ and $$BAA_1=30°$$. $$Let D\in [AB]$$ s.t. $$CD=AB$$.

I have to show that $$AB=AA_1$$.

This conclusion seems to be wrong because I can draw a triangle with $$AB>AA_1$$ but I am not sure if the circle with center $$C$$ and radius $$AB$$ intersects the edge $$[AB]$$.

• You're right - something's definitely wrong with the problem statement. – metamorphy Jan 4 at 2:03