Geometry inequality proof 
I started off with the given and by using the triangle inequality theorem but I don't know what to do next. Can someone please help? Thank you very much. I greatly appreciate it!
 A: $\angle ADC>\angle ACD$ and, because are both on first quadrant, then $\sin(ADC)>\sin(ACD)$.  Finally, using the sin rule, then $AD<AC$
A: Let's try the triangle inequality.
We observe that the length 
$|AS| + |AD| = |SD|$ (*)
because $|AS|$ and $|AD|$ are on the same line.
However, 
$|AS| + |AC| > |SC| = |SD|$ (**)
due to the triangle inequality. The strict inequality is because $|AS|$ and $|AC|$ do not lie in the same line. (Verify this, please). 
We have 
$|AD| = |SD| - |AS|$ by (*)
and 
$|AC| > |SD| - |AS| = |AD|$ by (**).
So $|AC| > |AD|$ as desired. 
A: You have $SA+AD=SD=SC$.
Also, because the sum of the lengths of two sides of the triangle is bigger than the length of the third side you have $SA+AC>SC$.
So you have $SA+AC>SA+AD$ from which follows $AC>AD$.
A: Since $\overline{SD}=\overline{SC}$, $\triangle SDC$ is isosceles, and therefore, $L$, the locus of points that are equidistant from $C$ and $D$, is the line through $S$ perpendicular to $\overline{CD}$. Since $A$ is on the same side of $L$ as $D$, $A$ is closer to $D$ than to $C$. That is,
$$
\overline{AD}\lt\overline{AC}
$$
