Question

m∠1 = 110°, m∠2 = 70°, ∠1 is supplementary to ∠3, ∠2 is supplementary to ∠3, prove ∠1 ≅ ∠2, copy the proof, reasons: definition of supplementary angles, linear pair theorem, copy and paste reasons, given, transitive property of congruence

Explanation:

Step1: Identify Given Angles

We know \( m\angle1 = 110^\circ \) and \( m\angle2 = 70^\circ \), and both \( \angle1 \) and \( \angle2 \) are supplementary to \( \angle3 \) (from the diagram's logic structure).

Step2: Recall Supplementary Angle Definition

Supplementary angles sum to \( 180^\circ \). For \( \angle1 \) and \( \angle3 \): \( m\angle1 + m\angle3 = 180^\circ \), so \( m\angle3 = 180^\circ - 110^\circ = 70^\circ \). For \( \angle2 \) and \( \angle3 \): \( m\angle2 + m\angle3 = 180^\circ \), so \( m\angle3 = 180^\circ - 70^\circ = 110^\circ \)? Wait, no—wait, the diagram shows \( \angle1 \cong \angle2 \)? Wait, no, the goal is to prove \( \angle1 \cong \angle2 \) using transitive property. Since \( \angle1 \) and \( \angle3 \) are supplementary (\( m\angle1 + m\angle3 = 180^\circ \)), \( \angle2 \) and \( \angle3 \) are supplementary (\( m\angle2 + m\angle3 = 180^\circ \)). By the congruent supplements theorem (if two angles are supplementary to the same angle, they are congruent), so \( \angle1 \cong \angle2 \). Also, using transitive property: if \( m\angle1 + m\angle3 = 180 \) and \( m\angle2 + m\angle3 = 180 \), then \( m\angle1 = m\angle2 \) (subtracting \( m\angle3 \) from both equations), so \( \angle1 \cong \angle2 \) by definition of congruent angles (equal measures).

Step3: Complete the Proof Logic

The given angles are \( m\angle1 = 110^\circ \), \( m\angle2 = 70^\circ \) (wait, no, maybe the diagram has \( \angle1 \) and \( \angle2 \) supplementary to \( \angle3 \)). Wait, the key is: \( \angle1 \) is supplementary to \( \angle3 \) (given or by linear pair), \( \angle2 \) is supplementary to \( \angle3 \) (given or by linear pair). Then by transitive property of equality (since \( m\angle1 + m\angle3 = 180 \) and \( m\angle2 + m\angle3 = 180 \), so \( m\angle1 = m\angle2 \)), hence \( \angle1 \cong \angle2 \).

Answer:

To prove \( \angle1 \cong \angle2 \):

1. Given/Definition: \( \angle1 \) is supplementary to \( \angle3 \) (\( m\angle1 + m\angle3 = 180^\circ \)), \( \angle2 \) is supplementary to \( \angle3 \) (\( m\angle2 + m\angle3 = 180^\circ \)).
2. Transitive Property of Equality: From \( m\angle1 + m\angle3 = m\angle2 + m\angle3 \), subtract \( m\angle3 \) from both sides: \( m\angle1 = m\angle2 \).
3. Definition of Congruent Angles: Angles with equal measures are congruent, so \( \angle1 \cong \angle2 \).

(If filling the diagram: the "linear pair postulate" justifies supplementary, "given" for angle measures, "congruent supplements theorem" or "transitive property" for \( \angle1 \cong \angle2 \).)

Final answer (for proving \( \angle1 \cong \angle2 \)): \( \angle1 \cong \angle2 \) by transitive property (or congruent supplements theorem) since both are supplementary to \( \angle3 \).