<Tri-2023-2063>, hint

Let <BCD=x, in triangle BDC BD/ sinBCD=CD/ sinCBD

(c/2).sinx=(bcosC/cosC.    sinx=1/2 (b=c) x=30.


Take $D'$ reflection of $D$ across $BC; \triangle BD'D\sim\triangle ABC$, their similarity ratio is $\frac{1}2$, i.e. $2DD'=BC=2CD=2CD'$, thence $\triangle CDD'$ is equilateral.

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