Question

prove that $\triangle mnq \cong \triangle pnq$.

(image of triangle with angles: at n, $\angle mnq = 30^\circ$, $\angle pnq = 30^\circ$; at q, $\angle mqn = 107^\circ$, $\angle pqn = 107^\circ$)

statement	reason
2. $m\angle mqn = m\angle \underline{\text{pick angle}} = 107^\circ$	given
3. $\underline{\text{pick statement}}$	theyre lengths of the same segment.
4. $\triangle mnq \cong \triangle pnq$	$\underline{\text{pick congruence criterion}}$ congruence

Explanation:

Step 1: Determine the equal angle for step 2

From the diagram, we see that \( m\angle MQN = m\angle PQN = 107^\circ \). So the angle to pick is \( \angle PQN \).

Step 2: Determine the equal side for step 3

The side \( NQ \) is common to both \( \triangle MNQ \) and \( \triangle PNQ \), so the statement is \( NQ = NQ \) (since they are lengths of the same segment).

Step 3: Determine the congruence criterion for step 4

We have two angles and the included side equal: \( \angle MNQ=\angle PNQ = 30^\circ \), \( NQ = NQ \), and \( \angle MQN=\angle PQN = 107^\circ \). This satisfies the Angle - Side - Angle (ASA) congruence criterion.

Answer:

1. \( m\angle MQN = m\angle PQN=107^\circ \) (Pick angle: \( \angle PQN \))
2. \( NQ = NQ \) (Pick statement: \( NQ = NQ \))
3. Congruence criterion: ASA (Angle - Side - Angle)