Prove that the sum of the three angles of a triangle is $180^{\circ}$.

(N/A) Let us consider a triangle $PQR$ where $\angle 1, \angle 2$,and $\angle 3$ are the interior angles of $\Delta PQR$.
We need to prove that $\angle 1 + \angle 2 + \angle 3 = 180^{\circ}$.
Draw a line $XPY$ parallel to $QR$ passing through the vertex $P$.
Since $XPY$ is a straight line,the sum of angles on it at point $P$ is $180^{\circ}$,so $\angle 4 + \angle 1 + \angle 5 = 180^{\circ}$.
Since $XPY \parallel QR$ and $PQ, PR$ are transversals,the alternate interior angles are equal:
$\angle 4 = \angle 2$ and $\angle 5 = \angle 3$.
Substituting these values into the equation $\angle 4 + \angle 1 + \angle 5 = 180^{\circ}$,we get:
$\angle 2 + \angle 1 + \angle 3 = 180^{\circ}$,which means $\angle 1 + \angle 2 + \angle 3 = 180^{\circ}$.
Thus,the sum of the three angles of a triangle is $180^{\circ}$.