Proof involving angle bisector in an arbitrary triangle

Question

In the above figure, AD is a bisector angle A (angle BAC). How do I prove in a triangle ABC of any dimensions that,

2. $AB > BD$
3. $AC > CD$

Is it also possible to prove that,

1. $AB > AD$
2. $AC > AD$

Answer 1

I think the first two can be proven. Let $BC=a, AC=b, AB=c$

Now, using angle bisector theorem,$$\frac{CD}{BD}=\frac{b}{c}$$ $$BD=\frac{c}{b+c}.a$$$$CD=\frac{b}{b+c}.a$$ But $b+c>a$ $$\implies BD<c$$$$\implies CD<b$$

However as mentioned the comments, the next two parts do not always hold.