Given: ΔABC Prove: m∠ZAB = m∠ACB + m∠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∠ZAB + m∠CAB = 180° by the. We also know m∠CAB + m∠ACB + m∠CBA = 180° because. Using substitution, we have m∠ZAB + m∠CAB = m∠CAB + m∠ACB + m∠CBA. Therefore, we conclude m∠ZAB = m∠ACB + m∠CBA using the.

A: angle addition postulate

B: of the triangle sum theorem

C. subtraction property

Just did the assignment

Linear Pairs; triangle angle sum property; equality and subtraction property.

Step-by-step explanation:

The first statement says that m∠ZAB + m∠CAB = 180°. These two angles are formed as linear pairs and from the Linear Pair postulate we know that if two angles form a linear pair, then the sum of their measures is 180° ... (1)

And 2nd statement stats m∠CAB + m∠ACB + m∠CBA = 180° which are angles inside the triangle. And from Angle sum property we know that the sum of the measures of the angles of a triangle is 180° ... (2)

From equation (1) and (2) we can equate m∠ZAB + m∠CAB = m∠CAB + m∠ACB + m∠CBA.

Now, using subtraction property we will subtract m∠CAB from both the sides and hence, we get m∠ZAB = m∠ACB + m∠CBA which is our desire result.