8.8 Mathematical Tutorial

\(\underline{\text{Proof: }}\) the sum of a triangle equals \(180°\)

Consider a triangle with vertices \(A\), \(B\) and \(C\)

In addition, denote \(\angle ABC=\beta,\angle ACB=\alpha\)

We need to show that \(\angle BAC=180^{\circ}-\alpha-\beta\)

Consider a segment \(DE\) which passes through vertex \(A\) and and parallel to \(BC\)

\(\angle DAB=\angle ABC=\beta\), due to the fact that they alternate between parallel segments \(DE//BC\)

Using the same idea we may also say that \(\angle EAC=\angle ACB=\alpha\)

Since \(\angle DAB+ \angle BAC+ \angle EAC= 180^{\circ }\) (straight angle), then (due to our notations) we get

\[ \beta+\angle BAC+\alpha=180^\circ \]

Resulting in: \(\angle BAC=180^{\circ}-\alpha-\beta\)

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