Final answer:
In a parallelogram ABCD, extending the sides AB and AD to points M and N respectively such that BM = AD and DN = AB, it can be proven that triangles BMC and DNC are isosceles. It can also be proven that points M, C, and N are collinear.
Step-by-step explanation:
In a parallelogram ABCD, extend the sides AB and AD to points M and N respectively such that BM = AD and DN = AB.
1°) Proving triangles BMC and DNC are isosceles:
Since AB || DC, and BM = AD, by alternate interior angles, we can conclude that angles BMC and CDA are congruent.
Similarly, since AD || BC, and DN = AB, we can conclude that angles DNC and BCA are congruent.
Therefore, in triangles BMC and DNC, we have angles BMC = CDA and angles DNC = BCA, which implies that both triangles are isosceles.
2°) Proving points M, C, and N are collinear:
Since AB || DC and the opposite sides of a parallelogram are parallel, we can conclude that line MN is parallel to AB and DC.
Since BM = AD and DN = AB, we can conclude that line MN is congruent to AD.
Therefore, points M, C, and N lie on the same line, making them collinear.