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Question
given: \\( \triangle abc \\) prove: \\( m\angle zab=m\angle acb + m\angle cba \\) we start with triangle \\( abc \\) and see that angle \\( zab \\) is an exterior angle created by the extension of side \\( ac \\). angles \\( zab \\) and \\( cab \\) are a linear pair by definition. we know that \\( m\angle zab + m\angle cab = 180^{circ} \\) by the we also know \\( m\angle cab + m\angle acb + m\angle cba = 180^{circ} \\) because using substitution, we have \\( m\angle zab + m\angle cab = m\angle cab + m\angle acb + m\angle cba \\). therefore, we conclude \\( m\angle zab = m\angle acb + m\angle cba \\) using the
Step1: Linear Pair Angles
Since angles \( \angle ZAB\) and \( \angle CAB\) are a linear pair, by the linear pair postulate \( m\angle ZAB + m\angle CAB=180^{\circ}\).
Step2: Triangle Angle - Sum
For \(\triangle ABC\), by the triangle angle - sum theorem \(m\angle CAB + m\angle ACB + m\angle CBA = 180^{\circ}\).
Step3: Subtraction Property
We have \(m\angle ZAB + m\angle CAB=m\angle CAB + m\angle ACB + m\angle CBA\). Subtract \(m\angle CAB\) from both sides. Using the subtraction property of equality (\(a + b=a + c\Rightarrow b = c\) when we subtract \(a\) from both sides of the equation \(a + b=a + c\)), we get \(m\angle ZAB=m\angle ACB + m\angle CBA\).
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The first blank is filled with "linear pair postulate", the second blank is filled with "triangle angle - sum theorem", and the last blank is filled with "subtraction property".