Question

given: \\( \triangle abc \\) prove: \\( m \angle zab = m \angle acb + m \angle cba \\) we start with triangle abc and see that angle \\( \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

Explanation:

Step1: Linear Pair Postulate

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 Theorem

For \(\triangle ABC\), by the Triangle Angle - Sum Theorem, \(m\angle CAB + m\angle ACB + m\angle CBA = 180^{\circ}\).

Step3: Subtraction Property of Equality

We have \(m\angle ZAB + m\angle CAB=m\angle CAB + m\angle ACB + m\angle CBA\). Subtracting \(m\angle CAB\) from both sides (using the Subtraction Property of Equality), we get \(m\angle ZAB=m\angle ACB + m\angle CBA\).

Answer:

The first blank: Linear Pair Postulate; the second blank: Triangle Angle - Sum Theorem; the third blank: Subtraction Property of Equality.