Introduction
One of the most elementary concepts we have learned about triangles in Geometry is the angle sum theorem. The theorem states that the sum the three interior angles of a triangle is $latex 180^{o}$. We can easily see this by duplicating or cutting the corners of a triangle and meeting the angles at a particular point (see first figure). The adjacent angles will form a straight angle which is equal to $latex 180^{o}$.
The proof of the angle sum theorem is quite easy. We just need to draw an extra line.
Given
Triangle $latex ABC$.
Theorem
The interior angle sum of triangle $latex ABC$ is equal to 180˚.
Proof
Draw a line segment passing through $latex A$ and parallel to $latex BC$. This is possible because given a line and a point, you can draw exactly one line through that point parallel to the first line.
Construct points $latex D$ and $latex E$ to the left and right of $latex A$ respectively as shown in the figure below. Now $latex BC$ and $latex DE$ are parallel, so $latex AB$ and $latex AC$ can be considered as transversals.
Now,
$latex \angle BAD \cong \angle ABC$ since they are alternate interior angles.
Also, $latex \angle EAC \cong \angle BCA$ since they are alternate interior angles.
$latex \angle BAD$, $latex \angle BAC$, and $latex \angle EAC$ form a straight angle therefore their sum is $latex 180^{o}$ .
But these three angles are also the interior angles of triangle $latex ABC$ (Can you see why?)
Therefore, the interior angle sum of triangle $latex ABC$ is $latex 180^{o}$.
Pingback: Triangle Angle Sum Theorem: A More Friendly Proof
Pingback: Proof of Angle Sum of Quadrilaterals
Pingback: Proof of the Polygon Angle Sum Theorem
Pingback: The Exterior Angle Theorem - Proofs from The Book