Question

rectangle mnop is shown below. aidan argues that δmnq ≅ δpoq.
what triangle congruence statements prove that the triangles are congruent?
sss, sas, aas, asa, hl
blank and blank

Explanation:

Step1: Analyze Rectangle Properties

In rectangle \( MNOP \), \( MN = PO \) (opposite sides of rectangle), \( \angle N=\angle P = 90^\circ \).

Step2: Find Angles in Triangles

In \( \triangle MNQ \), \( \angle N = 90^\circ \), \( \angle NQM = 40^\circ \), so \( \angle NMQ=180^\circ - 90^\circ - 40^\circ = 50^\circ \).
In \( \triangle POQ \), \( \angle P = 90^\circ \), \( \angle PQO=180^\circ - 100^\circ - 40^\circ = 40^\circ \)? Wait, no, \( \angle MQO = 100^\circ \), so \( \angle PQO = 180^\circ - 100^\circ - 40^\circ = 40^\circ \)? Wait, actually, \( \angle NQM = 40^\circ \), \( \angle MQO = 100^\circ \), so \( \angle PQO = 180^\circ - 100^\circ - 40^\circ = 40^\circ \)? Wait, no, linear pair: \( \angle NQM + \angle MQO + \angle PQO = 180^\circ \), so \( 40^\circ + 100^\circ + \angle PQO = 180^\circ \), so \( \angle PQO = 40^\circ \). So \( \angle NQM=\angle PQO = 40^\circ \), \( MN = PO \), \( \angle N=\angle P = 90^\circ \). So by AAS (two angles and a non - included side) or ASA? Wait, \( \angle N=\angle P = 90^\circ \), \( MN = PO \), \( \angle NQM=\angle PQO = 40^\circ \). So ASA: \( \angle N=\angle P \), \( MN = PO \), \( \angle NQM=\angle PQO \). Also, AAS: \( \angle N=\angle P \), \( \angle NMQ=\angle POQ \) (since \( \angle NMQ = 50^\circ \), \( \angle POQ=180 - 90 - 40 = 50^\circ \)), and \( MN = PO \). So AAS and ASA can be used. But let's check the given options. The triangle congruence criteria that apply: AAS (two angles and a non - included side) and ASA (two angles and included side). Wait, \( MN = PO \) (opposite sides of rectangle), \( \angle N=\angle P = 90^\circ \), \( \angle NQM=\angle PQO = 40^\circ \). So ASA (angle - side - angle: \( \angle N \), \( MN \), \( \angle NQM \) and \( \angle P \), \( PO \), \( \angle PQO \)) and AAS (angle - angle - side: \( \angle N \), \( \angle NMQ \), \( MN \) and \( \angle P \), \( \angle POQ \), \( PO \)). Also, since \( MN = PO \), \( \angle N=\angle P = 90^\circ \), \( \angle NQM=\angle PQO \), so ASA and AAS. But let's confirm the sides. In rectangle, \( MN = PO \), \( \angle N=\angle P = 90^\circ \), \( \angle NQM=\angle PQO \). So ASA (angle - side - angle) and AAS (angle - angle - side) are valid.

Answer:

AAS and ASA