Please find and open the attachment. We need to prove that all angles of a triangle, add upto [tex] 180\degree [/tex]. For this we will have to take the help of the diagram in the attachment.
We construct a straight line DE passing through B and parallel to the base AC of the triangle.
Now, we know that the measure of the angle of a straight line is [tex] 180\degree [/tex]. So, [tex] m\measuredangle DBA+m\measuredangle ABC+m\measuredangle CBE=180\degree [/tex].
Now, we know that since [tex] DE\parallel AC [/tex], then by the concept of "Alternate Interior Angles", [tex] m\measuredangle DBA=m\measuredangle BAC [/tex] and [tex] m\measuredangle EBC=m\measuredangle BCA [/tex].
If we now take the information of the above two paragraphs we come to the conclusion that we can substitute [tex] \angle BAC [/tex] for [tex] \angle DBA [/tex] and likewise, we can substitute [tex] \angle BCA [/tex] for [tex] \angle EBC [/tex].
Thus, [tex] m\measuredangle DBA+m\measuredangle ABC+m\measuredangle CBE=180\degree [/tex] will become
[tex] m\measuredangle BAC+m\measuredangle ABC+m\measuredangle BCA=180\degree [/tex]
thus proving that the interior angles of a triangle add up to a sum of [tex] 180\degree [/tex] because the angles [tex] \angle BAC [/tex], [tex] \angle ABC [/tex] and [tex] \angle BCA [/tex] are the interior angles of the [tex] \triangle ABC [/tex].
