Use of triangle inequality for 2 triangles

Question

In what angles of $\hat a, \hat b, \hat c, \hat d$ below conditions hold

1) $|AC+AD|>|BC+BD|$ or

1) $|AC+AD|<|BC+BD|$

Answer 1

Using the law of sines in triangle ABC we have $$\frac{AC}{\sin b}=\frac{AB}{\sin c}=\frac{BC}{\sin(b+c)},$$ where we have used that the angle at A in triangle ABC is $180-b-c$, and the fact that $\sin(180-x)=\sin x.$ Similarly applying the law of sines to triangle ABD we have $$\frac{BD}{\sin a}=\frac{AB}{\sin d}=\frac{AD}{\sin(a+d)}.$$ This gives the four desired lengths all in terms of the angles and $AB$ as $$AC=AB\frac{\sin b}{\sin c},\\ BC=AB\frac{\sin(b+c)}{\sin c},\\ BD=AB \frac{\sin a}{\sin d},\\ AD=AB\frac{\sin(a+d)}{\sin d}.$$ We now may substitute these into the inequality $AC+AD>BC+BD$ and cancel the common (positive) factor $AB$, to obtain the angle criterion $$\frac{\sin b}{\sin c}+\frac{\sin(a+d)}{\sin d} > \frac{\sin(b+c)}{\sin c}+\frac{\sin a}{\sin d}.$$ This cannot be simplified much, apart from maybe giving names to the third angles to the two triangles, or clearing denominators, etc. Of course the other inequality $AC+AD<BC+BD$ is equivalent to the above with the inequality reversed.

Side note: There is equality iff the two points $A,B$ lie on the same ellipse with foci at $C,D.$