Let ABC be an equilateral triangle with point D on AC.
Create a new triangle with side lengths AD, DC, BD.
Prove one angle of this new triangle is 120 degrees.
Place a copy of ABC, by rotating 60 degrees about vertex B.
B---------E
/*\** /
/ * \ ** /
/ * \ *F
/ * \ /
A--D------C
From the construction of this figure we have AD=CF, DC=FE, BD=BF, angle DBF=60, and angle DCF=120.
Now draw segment DF. This is the new triangle, and so far we already established side DC (itself), side CF (=AD) and angle DCF (=120 degrees). All we need is to find the distance of DF.
BD=BF and angle DBF is the subtending angle, then triangle DBF is isosceles. We also know that angle DBF=60, which makes triangle DBF an equilateral triangle. Therefore BD=DF.
Then this means we have constructed triangle DCF as a triangle with sides congruent to AD, DC, and BD and DCF is a 120 degree angle, satisfying the problem.
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