Unfortunately, the answer of user35593 is broken but I obtained something in the same line : $$\cos(B(t)C(t)) = \cos(tAB-tAC) - \sin(tAB)\sin(tAC)(1-\cos a),$$ where $a$ is the angle at $A$. Thus $$\cos(B(t)C(t)) \geq \cos(AB - AC) + \cos a - 1 = \sin(AC) + \cos a -1.$$ We assume that $BC \leq \frac\pi2$, so that $\cos a \sin(AC) = \cos(BC) \geq 0$. This is enough to have $$\sin(AC) + \cos a -1 \geq 2 \cos(a)^2 \sin(AC)^2 - 1 = \cos(2 BC).$$ This gives the claim.

Unfortunately, the answer of user35593 is broken but I obtained something in the same line : $$\cos(B(t)C(t)) = \cos(tAB-tAC) - \sin(tAB)\sin(tAC)(1-\cos a),$$ where $a$ is the angle at $A$. Thus $$\cos(B(t)C(t)) \geq \cos(AB - AC) + \cos a - 1 = \sin(AC) + \cos a -1.$$ We assume that $BC < \frac\pi2$, so that $\cos a \sin(AC) = \cos(BC) > 0$. Since $0\leq AC \leq \pi$ we have $\sin(AC)\geq 0$ and hence $\sin(AC)>0$ and $\cos(a)>0$. This is enough to have $$\sin(AC) + \cos a -1 \geq 2 \cos(a)^2 \sin(AC)^2 - 1 = \cos(2 BC).$$ This gives the claim.