The proof that ΔMNS ≅ ΔQNS is shown. Given: ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. Prove: ΔMNS ≅ ΔQNS

We know that ΔMNQ is isosceles with base MQ. So, MN ≅ QN by the definition of isosceles triangle. The base angles of the isosceles triangle, ∠NMS and ∠NQS, are congruent by the isosceles triangle theorem. It is also given that NR and MQ bisect each other at S. Segments_____are therefore congruent by the definition of bisector. Thus, ΔMNS ≅ ΔQNS by SAS.

NS and NS
NS and RS
MS and RS
MS and QS

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SerenaBochenek SerenaBochenek · Answer 1 · 2019-03-11T05:29:25+01:00

Answer:

The correct option is MS and QS

Step-by-step explanation:

Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.

As NR and MQ bisect each other at S

⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ

In ΔMNS and ΔQNS

MN=QN (∵ MNQ is isosceles triangle)

∠NMS=∠NQS (∵ MNQ is isosceles triangle)

MS=SQ (Given)

By SAS rule, ΔMNS ≅ ΔQNS.

Hence, segments MS and SQ are therefore congruent by the definition of bisector.

The correct option is MS and QS

MrRoyal MrRoyal · Answer 2 · 2021-10-06T10:54:01+02:00

When a point bisects a line, the point divides the line into equal halves

The option that complete the proof is: (d) MS and QS

From the attached figure, we have the following observations

The above highlights mean that:

From the question, the triangles whose congruence are to be proved are:

[tex]\mathbf{\triangle MNS\ and\ \triangle QNS}[/tex]

Already, we have that:

[tex]\mathbf{\angle NMS \cong \angle NQS}[/tex]

Since the angles at M and Q are congruent, the length MS must also be congruent to the length QS

Hence, the statement that completes the proof is:

[tex]\mathbf{MS \cong QS}[/tex].

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