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Question
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 ______ 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
Step1: Recall the definition of bisect
If two segments bisect each other, the point of intersection divides each segment into two equal - length parts.
Since \(\overline{NR}\) and \(\overline{MQ}\) bisect each other at \(S\), for segment \(\overline{MQ}\), \(S\) is the mid - point.
Step2: Use the mid - point property
By the definition of a mid - point (a point that bisects a segment), if \(S\) is the mid - point of \(\overline{MQ}\), then \(MS = QS\).
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MS and QS