Question

1. given: ( overline{am} cong overline{cp}, overline{cm} cong overline{gp} ), c is the midpoint of ( overline{ag} ) prove: ( \triangle acm cong \triangle cgp )

Explanation:

Step1: Definition of midpoint

Since \(C\) is the midpoint of \(\overline{AG}\), by the definition of a midpoint, \(AC = CG\).

Step2: Given information

We are given that \(\overline{AM}\cong\overline{CP}\), so \(AM = CP\).

Step3: Given information

We are given that \(\overline{CM}\cong\overline{GP}\), so \(CM = GP\).

Step4: SSS (Side - Side - Side) congruence criterion

In \(\triangle ACM\) and \(\triangle CGP\), we have \(AC = CG\) (from Step 1), \(AM = CP\) (from Step 2), and \(CM = GP\) (from Step 3).

Step5: Conclusion

By the SSS (Side - Side - Side) congruence criterion, \(\triangle ACM\cong\triangle CGP\).

Answer:

Statements	Reasons
2. \(AC = CG\)	2. Definition of midpoint
3. \(AM = CP\)	3. Given (\(\overline{AM}\cong\overline{CP}\))
4. \(CM = GP\)	4. Given (\(\overline{CM}\cong\overline{GP}\))
5. \(\triangle ACM\cong\triangle CGP\)	5. SSS (Side - Side - Side) congruence criterion