Question

what theorem proves that \\( \triangle nvp \cong \triangle vmq \\)?

angle angle side
angle side angle
side angle side
hypotenuse leg

Explanation:

Step1: Analyze the given triangles

We have two triangles \(\triangle NVP\) and \(\triangle MVQ\). We know that \(\angle NVP=\angle MVQ\) (vertically - opposite angles). Also, \(NP = MQ\) (given as marked equal) and \(\angle NPV=\angle MQV = 90^{\circ}\) (right - angles).

Step2: Recall the congruence theorems

• Angle - Angle - Side (AAS): If two angles and a non - included side of one triangle are equal to two angles and the corresponding non - included side of another triangle, then the two triangles are congruent.
• Angle - Side - Angle (ASA): If two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the two triangles are congruent.
• Side - Angle - Side (SAS): If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent.
• Hypotenuse - Leg (HL): This is for right - angled triangles. If the hypotenuse and a leg of one right - angled triangle are equal to the hypotenuse and a leg of another right - angled triangle, then the two right - angled triangles are congruent.

In our case, we have two right - angled triangles (\(\angle NPV=\angle MQV = 90^{\circ}\)), a pair of equal angles (\(\angle NVP=\angle MVQ\)) and a pair of equal non - hypotenuse sides (\(NP = MQ\)).

Answer:

Angle - Angle - Side (AAS)