Question: In triangle $ABC$, $D$ lies on side $BC$ such that $\overline{AD}$ bisects $\angle BAC$ and $\angle ADB$ is obtuse.
Which of the following could be true? (There might be more than one correct answer.)
(A) $AD<BC$ and $CD \geq AB$
(B) $AD \geq BC$ and $CD<AB$
(C) $AD<BC$ and $CD<AB$
(D) $AD \geq BC$ and $CD \geq AB$
Here's the observations I made:
Using the angle bisector theorem, if $AC=a$ and $AB=b$, then we can say that $CD=ax$ and $BD=bx$.
In $\triangle ABD$, because $AB>DB$, it follows that $0<x<1$.
Using this fact in $\triangle ACD$, it follows that $\\angle CAD=\angle DAB>180^{\circ}-\angle ADB$.
Returning back to the original question, my idea was to consider the shared sides between the pairs of triangles relevant to the lengths mentioned in the options. However, all the work I've done with this has resulted in a dead end.
