Question

given:
rm = sn
tm = tn
prove:
rn = sm

1. given rm = sn, tm = tn
2. reflexive rm + tm = sn + tn
3. addition property of equality rn = sm
4. betweeness rt = st
5. substitution rm + tm = rt, sn + tn = st
6. sas ∠t = ∠t
7. cpcte triangle rtn congruent to triangle stm

Explanation:

Step1: Match the given with the reason

Given \(RM = SN\) and \(TM=TN\) which is the first step.

Step2: Use reflexive property

\(\angle T=\angle T\) (reflexive property of angles, an angle is congruent to itself).

Step3: Use addition property of equality

Since \(RM = SN\) and \(TM = TN\), by addition property of equality \(RM+TM=SN + TN\).

Step4: Use betweeness

If \(R - M - T\) and \(S - N - T\), then \(RM+TM = RT\) and \(SN+TN=ST\) (segment addition postulate: if a point \(M\) is between \(R\) and \(T\), then \(RM + MT=RT\)).

Step5: Use substitution

Substitute \(RM + TM\) with \(RT\) and \(SN+TN\) with \(ST\) in \(RM + TM=SN + TN\), we get \(RT = ST\).

Step6: Use SAS (Side - Angle - Side)

In \(\triangle RTN\) and \(\triangle STM\), \(RT = ST\) (from step 5), \(\angle T=\angle T\) (step 2), \(TN=TM\) (given). So, \(\triangle RTN\cong\triangle STM\) (SAS congruence criterion).

Step7: Use CPCTE (Corresponding Parts of Congruent Triangles are Equal)

Since \(\triangle RTN\cong\triangle STM\), then \(RN = SM\) (CPCTE).

Answer:

1. \(RM = SN, TM = TN\)
2. \(\angle T=\angle T\)
3. \(RM + TM=SN + TN\)
4. \(RM + TM = RT, SN + TN=ST\)
5. \(RT = ST\)
6. \(\triangle RTN\cong\triangle STM\)
7. \(RN = SM\)