**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 $latex 180^{o}$.

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