Question

Given: C∉ BD ,△ABC D∈ ray BC ,AB=AC=BC Prove: BD>DA>AB

Answer 1

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,

$AD^2=AC^2+CD^2-2\cdot AC\cdot CD\cdot \cos \angle ACD,\\\\AD^2=a^2+x^2-2\cdot a\cdot x\cdot \cos 120^{\circ},\\\\AD^2=a^2+x^2+ax,\\\\AD=\sqrt{a^2+x^2+ax}.$

Since $a^2+x^2+ax=a^2+x^2+ax+ax-ax=(a+x)^2-ax,$ then

$AD^2=(a+x)^2-ax$

and

$AD^2=a^2+x^2+ax>a^2=AB^2\Rightarrow AD>AB.$

Therefore, you get double inequality

$AB$ or $BD>AD>AB.$