the proof that $ \triangle mns \cong \triangle qns $ is shown. select the answer that best completes the proof.
given: $ \triangle mnq $ is isosceles with base $ \overline{mq} $, and $ \overline{nr} $ and $ \overline{mq} $ bisect each other at s.
prove: $ \triangle mns \cong \triangle qns $
we know that $ \triangle mnq $ is isosceles with base $ \overline{mq} $. so, $ \overline{mn} \cong \overline{qn} $ by the definition of isosceles triangle. the base angles of the isosceles triangle, $ \angle nms $ and $ \angle nqs $, are congruent by the isosceles triangle theorem. it is also given that $ \overline{nr} $ and $ \overline{mq} $ bisect each other at s.
segments $ \underline{\quad\quad} $ are therefore congruent by the definition of bisector. thus, $ \triangle mns \cong \triangle qns $ by sas.
$ \bigcirc $ ns and qs
$ \bigcirc $ ns and rs
$ \bigcirc $ ms and rs
$ \bigcirc $ ms and qs

Question

Accepted Answer

# Explanation:
## Step1: Recall the definition of bisector
If two segments bisect each other, the segments formed by the intersection are congruent.
Since $\overline{NR}$ and $\overline{MQ}$ bisect each other at $S$, for segment $\overline{MQ}$, $S$ is the mid - point.
## Step2: Apply the mid - point property
By the definition of a bisector (mid - point when bisecting a line segment), if $S$ bisects $\overline{MQ}$, then $MS = QS$.

# Answer:
MS and QS

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the definition of bisector

Step2: Apply the mid - point property

Answer: