Triangle $\Delta ABC$: Proof $\angle ADB<\angle ACB$ I have to proof the following statement:
Consider a triangle $\Delta ABC$ with point $D$ inside. Then $\angle ADB>\angle ACB$ holds.

I tried to draw a parallel line through C und D to AB, but I don't know how to go on. Any hints are welcome.
 A: Ah I think I got it ($\gamma$ is the angle in A, $\delta$ is the angle in B):
$\alpha=180°-\gamma_1-\delta_1$ and $\beta=180°-\gamma-\delta$, since $\gamma=\gamma_1+\gamma_2$ and $\delta=\delta_1+\delta_2$ we have $\beta=180°-(\gamma_1+\gamma_2)-(\delta_1+\delta_2)=180°-\gamma_1-\gamma_2-\delta_1-\delta_2$ which is obviously smaller then $\alpha$
Am I right?
A: As suggested by David Mitra: Look at the two triangles $\triangle ADB$ and $\triangle CAB$. We know the sum of the angles of each of these triangles is $180^\circ$. That means we can express $$\alpha = 180^\circ - \cdots - \cdots$$
$$\beta = 180^\circ - \ldots - \ldots$$

I just saw your "answer" post: 
Exactly: just clarify which angles $\gamma, \; \delta$ define (with respect to the larger triangle) and similarly, identify $\gamma_1$, $\gamma_2$, $\delta_1$ and $\delta_2$, in terms of which angles they correspond to with respect to, e.g., the unnamed angles of the interior triangle.  
And then, yes, you will have $\gamma=\gamma_1+\gamma_2$ and $\delta=\delta_1+\delta_2$ 
Nice work.
A: The simplest way would be to note that $\angle BAD < \angle BAC$ and $\angle ABD < \angle ABC$. Since $\angle BAD+\angle ABD+\angle ADB = \angle BAC+\angle ABC+\angle ACB$, we can bring the compared terms to the same side:
$(\angle ADB - \angle ACB) = (\angle BAC - \angle BAD) + (\angle ABC - \angle ABD)$
Both the terms $(\angle BAC - \angle BAD)$ and $(\angle ABC - \angle ABD)$ must be positive, because of our earlier  note. Then we know that $(\angle ADB - \angle ACB)$ must be positive as well, so we must have $\angle ADB > \angle ACB$.
