The proof that AMNS AQNS is shown. Given: AMNQ is isosceles with base MQ, and NR and MQ
bisect each other at S.
Prove: AMNS AQNS
We know that AMNQ is isosceles with base MQ. So,
MN - QN by the definition of isosceles triangle. The base
angles of the isosceles triangle, ZNMS and 2NQS, 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,
AMNS AQNS by SAS.
N
M
о
S
NS and QS
ONS and RS
MS and RS
MS and QS
R

The answer 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

d? maybe

step-by-step explanation:

false

step-by-step explanation:

false.   two legs of the trapezoid would be congruent, but the other two would not.   were all the legs congruent, then we'd have a square; were the 2 shorter sides congruent and the 2 longer sides also congruent, we'd have a rectangle.