Question

17 Extend respectively the sides [AB] and [AD] of a parallelogram ABCD such that BM = AD and DN = AB.

1°) Prove that the triangles BMC and DNC are isosceles. 2°) Prove that the points M, C and N are collinear​

Answer 1

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.