How should I prove that the two lines meet?

Question

In the triangle $\,ABD\,$: $$ \angle{ABD}=30^\circ\,,\quad AB=5.6\,\text{cm}\,,\quad DB=3\,\text{cm} $$ And the side $\,BD\,$ is produced to C.

NOW, if angle $\,\angle{DAX}\,$ is constructed such that $\,\angle{DAX}=\angle{ADC}\,$, THEN:
PROVE THAT $\,AX\,$ WILL INTERSECT $\,DC\,$ AT A POINT $\,E\,$?

For proving this, I think we need to show that $\,\angle{ADC}+\angle{DAX}\lt180^\circ\,$,
But I do not find any way of doing this. (I have wasted many hours). Please help.

Answer 1

Strategy: To show that $\angle ADC < 90 ^\circ$.

This is equivalent to showing that $\angle ADB > 90^\circ$.

Now we could proceed by using the cosine and sine theorems which can be used to calculate all lengths and angles in $\triangle BAD $.

However, as we do not need to find the exact value of $\angle ADB$, we proceed differently.

Let $\mathcal{l}$ be the line normal to the line $BD$ in the point $D$, and let $F$ be the point of intersection between $\mathcal{l}$ and the line (not necessarily the line segment) $BA$. Then $\triangle BFD$ is a $30-60-90$ triangle and hence we can calculate the length of the line segment $ \overline{BF}$.

Finally, compare the length of $ \overline{BF}$ to the length of $\overline{BA}$. What can you conclude about $\angle ADB$ ?