### Theory:

Now recall the angle sum property of a triangle that we studied earlier.
The sum of the measure of three angles of a triangle is $$180°$$.

Proof:

Consider a triangle $$ABC$$ with interior angles measures $$∠1$$, $$∠2$$ and $$∠3$$.

Draw a line $$DE$$ parallel to $$BC$$.

Now the angle formed by the parallel line $$DE$$ with the triangle $$ABC$$ is $$∠4$$ and $$∠5$$.

Since $$DE$$ is parallel to $$BC$$, using the alternate interior angle property $$∠2$$ must equal to $$∠4$$. Similarly, $$∠3$$ must be equal to $$∠5$$.

That is $$∠2 =∠4$$ and $$∠3=∠5$$. As $$DE$$ is a straight line, $$∠5$$ and $$∠CAD$$ are linear pairs (Pair of adjacent supplementary angles).

$$∠5 + ∠CAD = 180°$$ .

That is, $$∠5 + ∠1 + ∠4 = 180°$$.

Equivalently, $$∠1 + ∠2 + ∠3 = 180°$$.

It states that the total measures of the three angles of a triangle is $$180°$$.

Let's extend the angle sum property to the quadrilateral.
The sum of all the angles of a quadrilateral is $$360°$$.

Proof:

Let $$ABCD$$ be a quadrilateral.

Cut the quadrilateral into two triangles by drawing one of its diagonals $$AC$$.

It is obvious from the figure that $$∠1+∠2 =∠A$$ and $$∠3+∠4 =∠C$$.

We know that the sum of all the angles of a triangle is $$180°$$.

Consider the triangle $$ABC$$.

$$∠1 + ∠4 + ∠B =180°$$.

and $$∠2 + ∠3 + ∠D =180°$$.

Adding the above equations will result as follows:

$$∠1 + ∠4$$ $$+ ∠B$$ $$+ ∠2$$ $$+ ∠3$$ $$+ ∠D$$ $$=$$ $$180° + 180°$$.

Arranging the angles in order becomes:

$$(∠1 + ∠2)$$ $$+ ∠B +$$ $$(∠3 + ∠4)$$ $$+ ∠D$$ $$= 360°$$.

Substitute the known values $$∠1+∠2 =∠A$$ and $$∠3+∠4 =∠C$$,

$$∠A$$ $$+ ∠B$$ $$+ ∠C$$ $$+ ∠D$$ $$= 360°$$.

Thus, the sum of the angles of a quadrilateral is $$360°$$.