Triangle ABC is equilateral, because AB=BC=AC=a. Then
m∠A=m∠B=m∠C=60°.
Let point D lie on the ray BC to the right from points B and C and let CD=x. Then BD=a+x, AB=a.
Consider triangle ACD. In this triangle, m∠ACD=180°-m∠ACB=180°-60°=120°.
By the cosine theorem,
[tex]AD^2=AC^2+CD^2-2\cdot AC\cdot CD\cdot \cos \angle ACD,\\\\AD^2=a^2+x^2-2\cdot a\cdot x\cdot \cos 120^{\circ},\\\\AD^2=a^2+x^2+ax,\\\\AD=\sqrt{a^2+x^2+ax}.[/tex]
Since [tex]a^2+x^2+ax=a^2+x^2+ax+ax-ax=(a+x)^2-ax,[/tex] then
[tex]AD^2=(a+x)^2-ax<(a+x)^2=BD^2\Rightarrow AD<BD[/tex]
and
[tex]AD^2=a^2+x^2+ax>a^2=AB^2\Rightarrow AD>AB.[/tex]
Therefore, you get double inequality
[tex]AB<AD<BD[/tex] or [tex]BD>AD>AB.[/tex]
