The first part Let $BC=CD=a , AD=c , AB=b$ First using Cosine formula in triangle $ADC$, $AC^2=a^2+c^2-2acCos120=a^2+c^2 +ac$ Now again applying Cosine formula in triangle $ABC$ $AC^2=a^2+b^2-2abCos60$ $a^2+c^2+ac=a^2+b^2-ab$ That is $(b+c)(b-c-a)=0$ Thus the result. I think(just a thought as of now) converse is also true and can be obtained by working backwards.

Let E be midpoint of arc AC not containing D thus DE=DA+DC the equivalence is clear and the remaining of the proof from the diagram

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