$|BC|=|AD|$ and points $A,B,C,D$ lie in this order on one circle, so $|AC|>|CD|$. Hence there exists point $E$ on segmeny $AD$ such that $|CD|=|CE|$. Then triangle $BCE$ is equilateral and $\triangle ADC\equiv\triangle AEC$ by a-s-a: ray $AC^\rightarrow$ is bisector of angle $DAB$ and $\angle AEC=120^\circ=\angle ADC$. Thus $|AB|=|AE|+|BE|=|AD|+|CD|$.

Join BD Since ABCD cyclic, angle ABC+angle ADC=180° angle ADC=180°-60°=120° Let angle DBC=x, so angle angle BDC=x So, angle BCD=180°-2x therefore, angle ABD=60°-x angle ADB=120°-x In ∆ BCD, BY sine rule sin(180°-2x)/BD=sin x/ CD sin 2x/BD=sin x/ CD Since ABCD is cyclic, angle BAC= 180°- angle BCD= 2x In ∆ ABD, By sine rule , sin 2x/BD= sin( 60°-x)/AD= sin(120°-x)/AB sin 2x/BD=sin x/ CD SO, sin x/CD=sin (60°-x)/AD Now, by addendo, sin x/CD=sin (60°-x)/AD= [sin x+ sin(60°-x)]/CD+AD= sin(120°-x)/AB cos(x-30°)/CD+DA=sin (120°-x)/AB Since, cos(x-30°)= sin(120°-x) So, CD + DA = AB