Prove the sum of intern angles of a quadrilateral is equal to 360º

You can use the information: theorem "The sum of all angles in a triangle is equal to $180º$" to prove the sum of all angles in a quadrilateral is 360º.

I know considering two parallel lines, the altern intern angles theorem and the info that perpendicular lines form 4 angles of 90º would be useful for this proof, but I am not quite able to organize them and build the proof

Call the angles of the quadrilateral $A, B, C, D$, with $A$ opposite of $D$ and $B$ opposite of $C$. Our goal is to find $\angle A+\angle B+\angle C+\angle D$.

Create the line $BD$. We now have two triangles, $\triangle ABD$ and $\triangle CBD$.

We have

$$\angle B=\angle ABD+\angle CBD\quad\text{ and }\quad\angle D =\angle ADB+\angle CDB.$$

Moreover, since we have triangles, we know that

$$\angle A+\angle ABD+\angle ADB=180^\circ\quad\text{ and }\quad \angle C+\angle CBD+\angle CDB=180^\circ$$

Therefore,

\begin{align*} \angle A+\angle B+\angle C+\angle D&=\angle A+(\angle ABD+\angle CBD)+\angle C+(\angle ADB+\angle CDB)\\ &=[\angle A+\angle ABD+\angle ADB]+[\angle C+\angle CBD+\angle CDB]\\ &=180^\circ+180^\circ\\ &=360^\circ \end{align*}