Triangle Angle Sum Theorem

Given: In △ABC in the figure above,

• $\mathrm{\angle }\mathrm{BAC}$, $\mathrm{\angle }\mathrm{ABC}$ and $\mathrm{\angle }\mathrm{ACB}$ are interior angles.
• $\mathrm{\angle }\mathrm{BAC}=a$, $\mathrm{\angle }\mathrm{ABC}=b$ and $\mathrm{\angle }\mathrm{ACB}=c$

Construction:

• Drawn line segment PQ parallel to side BC passing through the vertex A.
• Angles $\mathrm{\angle }\mathrm{PAB}$ and $\mathrm{\angle }\mathrm{QAC}$ are formed by the line segment PQ.
• Consider $\mathrm{\angle }\mathrm{PAB}=x$ and $\mathrm{\angle }\mathrm{QAC}=y$

To Prove: We have to prove that the sum of the angles $\angle \mathrm{ABC}$, $\mathrm{\angle }\mathrm{BAC}$ and $\angle \mathrm{ACB}$ is 180°, i.e. $$\angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180°$$

Proof: Sides AB and AC are transversals for the parallel lines PQ and BC.
Hence, according to the alternate interior angles theorem, $$\begin{array}{}\mathrm{(1)}& x& =& b\mathrm{(2)}& y& =& c\end{array}$$ The sum of the angles that are formed on a straight line at the same point is always 180°.
Therefore the sum of the angles $x$, $a$ and $y$ is 180°, since these angles lie on the straight line PQ. $$\begin{array}{}\mathrm{(3)}& \therefore x+a+y=180°\end{array}$$ Put values from $\mathrm{(1)}$ and $\mathrm{(2)}$ in above equation $\mathrm{(3)}$. $$\begin{array}{r}\therefore b+a+c=180°\\ \therefore \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180°\end{array}$$ Henced proved the triangle angle sum theorem.