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To prove that the sum of the angles of a triangle is 180 degrees, take a general triangle ABC.
Draw a line PQ through the point B parallel to AC. The lines AB and CB are transversal lines through the parallel lines PQ and AC. The angles BAC and ABP and the angles CBQ and BCA are alternate interior angles and are equal.
The three angles PBA, ABC and CBQ lie around one side of the straight line PQ and their sum is equal to 180 degrees.
This proves that the sum of the angles of a triangle is equal to 180 degrees.
Let ABC be a triangle.
Exend AC to X from C.
Draw a line CD frm C which is || to AB.
Now AB || CD and BC is a transvalal.
Therefore angle ABC = angle BCD being alternate angle.
Therefore angle CAB = angle DCX being corresponding angles.
Therefore angle CAB + angle ABC +angle ACB = Angle ACB+Angle ABD+Angle DCX. The right side is the sum of angle on the straight line, which is 180 degree.
This proves that the sum of the angles of a triangle is 180 degrees.
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