Question

given: mq = nq; q is the midpoint of \\(\overline{lp}\\); \\(\overline{lm} \cong \overline{pn}\\) which congruence theorem can be used to prove \\(\triangle mlq \cong \triangle npq\\)? \\(\bigcirc\\) aas \\(\bigcirc\\) sss \\(\bigcirc\\) asa \\(\bigcirc\\) sas

Explanation:

Step1: Analyze the given information

We know that \(MQ = NQ\) (given), \(\overline{LM}\cong\overline{PN}\) (given), and since \(Q\) is the mid - point of \(\overline{LP}\), then \(LQ=PQ\).

Step2: Recall the SAS (Side - Angle - Side) congruence theorem

The SAS congruence theorem states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
In \(\triangle MLQ\) and \(\triangle NPQ\), we have:

• \(LM = PN\) (given)
• \(LQ = PQ\) (because \(Q\) is the mid - point of \(LP\))
• \(\angle MLQ\) and \(\angle NPQ\) are not relevant for SAS. But if we consider the vertical angles (not shown in the given information, but in the geometric figure, the angles between \(LM\) and \(LQ\) and \(PN\) and \(PQ\) with the sides \(MQ\) and \(NQ\) are not needed as we have two sides \(LM = PN\), \(LQ = PQ\) and \(MQ = NQ\) (the third side in the two triangles \(\triangle MLQ\) and \(\triangle NPQ\) which forms the SAS condition).

So, the SSS (Side - Side - Side) congruence theorem is not applicable here as we are not using three pairs of sides for the proof. AAS (Angle - Angle - Side) and ASA (Angle - Side - Angle) are not applicable as we are not using two pairs of angles. The SAS (Side - Side - Side) congruence theorem is used as we have two pairs of sides (\(LM = PN\), \(LQ = PQ\)) and the included side \(MQ = NQ\) (the side between the two pairs of sides in the two triangles \(\triangle MLQ\) and \(\triangle NPQ\)).

Answer:

D. SAS