<Tri-2026-2311>, hint

Let E on BC such that BE/EC=b/c.

Since AB/AC=ab/ac=b/c=bd/cd=BD/CD, A,D,E are collinear 

and ADE is angle bisector of <BAC and <BDC

Similarly BD is angle bisector of <ABC. 

<BDE=<A/2+<B/2, <CDE=<A/2 +<C/2, <BDE=<CDE ---> <B=<C

Hence ABC is equilateral  1/2(<B+<C)=60