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Midpoints of the Sides of a Paralellogram


Alignments to Content Standards: G-CO.C.11

Task

Suppose that $ABCD$ is a parallelogram, and that $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively. Prove that $MN = AD$, and that the line $\overleftrightarrow{MN}$ is parallel to $\overleftrightarrow{AD}$.

Gco.11_midpointsprob_057badd12ec7b383cf0c58505d38cea6

IM Commentary

This is a reasonably direct task aimed at having students use previously-derived results to learn new facts about parallelograms, as opposed to deriving them from first principles. The solution provided (among other possibilities) uses the SAS trial congruence theorem, and the fact that opposite sides of parallelograms are congruent.

Solution

Parallelogrammidpoin_ef4da8108c21a105640f6f7954341897

The diagram above consists of the given information, and one additional line segment, $\overline{MD}$, which we will use to demonstrate the result. We claim that triangles $\triangle AMD$ and $\triangle NDM$ are congruent by SAS:

  1. We have $\overline{MD}=\overline{DM}$ by reflexivity.
  2. We have $\angle AMD= \angle NDM$ since they are opposite interior angles of the transversal $MD$ through parallel lines $AB$ and $CD$.
  3. We have $\overline{MA}=\overline{ND}$, since $M$ and $N$ are midpoints of their respective sides, and opposite sides of parallograms are congruent: $$ \overline{MA}=\frac{1}{2}(\overline{AB})=\frac{1}{2}(\overline{CD})=\overline{ND}. $$
Now since corresponding parts of congruent triangles are congruent, we have $DA=NM$, as desired. Similarly, we have congruent opposite interior angles $\angle DMN\cong \angle MDA$, so $\overleftrightarrow{MN}$ is parallel to $\overleftrightarrow{AD}$.

Evan says:

about 1 year

Hi team, Trying to prove this with rigid motions (either translation or rotation of one half quad onto the other). Would love feedback on this approach - especially how to demonstrate that the line segments are parallel efficiently. Thanks in advance, Evan.

(1) Since parallelograms have 180 degree rotational symmetry, opposite sides of a parallelogram are both parallel and congruent, therefore: - AB || DC - DC = AB and AD = BC (2) M and N are midpoints of their respective sides, and opposite sides of parallelograms are congruent (1), therefore: - MA=ND=MB=NC

If we translate the quad MNCB along the parallel lines AB and DC (1) by a vector such that point N' is in the position of D then M' will be in the position of A since the distance translated will be equal, MA=ND (2) and the direction is equal, definition of parallel lines. By the same reasoning, point C' will be in the position of N and B' will be in the position of M. Let's call this translation T(quad).

(3) Since this rigid motion, T(quad), maps M'N'C'B' on to ADNM, we conclude that these two quadrilaterals are congruent. Therefore corresponding angles and sides of the two quads are congruent. - MN = AD

(4) Since a vector translation of MN maps directly on to AD without rotation in the plane, all distances between points along the vector between the two lines are equal, and those two lines are parallel. - MN || AD

wmukluk says:

over 5 years

step a. REFLEXIVITY!!!

Cam says:

over 5 years

Thanks!

Sometimes I think adding in extra syllables is just ... reflexive.