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Midpoints of Triangle Sides
Task
Suppose $ABC$ is a triangle. Let $M$ be the midpoint of $\overline{AB}$ and $P$ the midpoint of $\overline{BC}$ as pictured below:
 Show that $\overleftrightarrow{MP}$ and $\overleftrightarrow{AC}$ are parallel.
 Show that $AC = 2MP$.
IM Commentary
The goal of this task is to use similarity transformations to relate two triangles. The triangles in question are obtained by taking midpoints of two sides of a given triangle. In the picture above, $\triangle BAC$ can be seen as a scaled version of $\triangle BMP$, with scale factor 2 and center of dilation $B$. Equivalently, $\triangle BMP$ is the scaled version of $\triangle BAC$ with scale factor $\frac{1}{2}$ and center of dilation $B$. The task uses the important fact that a dilation maps a line $\ell$ not containing the center of dilation to a line parallel to $\ell$: this is verified experimentally in GSRT.1a and is an axiom in the transformational approach to geometry. Also important in this task is GSRT.1b which allows us to conclude that the dilation with scale factor 2 and center of dilation $B$ doubles the length of $\overline{MP}$.
In many traditional approaches to high school geometry, the result of this task is proven with the SAS similarity theorem: if two triangles share an angle and if the sides making the angle are proportional then the two triangles are similar. This is Book VI, Proposition 6 of Euclid and the proof is very technical, relying on Propositions 1, 2, and 4 of Book VI. The transformational approach to geometry replaces the SAS similarity theorem with properties of dilations: they preserve angle measures, scale all line segment lengths by the same (nonzero) factor, and take lines not passing through the center of dilation to parallel lines.
Solution

Suppose $D$ is the dilation with scale factor 2 and center $B$. Since $M$ is the midpoint of $\overline{AB}$, we know that $D(M) = A$. Since $P$ is the midpoint of $\overline{BC}$ we know that $D(P) = C$. Since dilations map lines to lines, we know that $D\left(\overleftrightarrow{MP}\right) = \overleftrightarrow{AC}$. Dilations take lines not containing the center of dilation to parallel lines so we can conclude that $\overleftrightarrow{MP}$ is parallel to $\overleftrightarrow{AC}$.
 Since $D\left(\overline{MP}\right) = \overline{AC}$ and $D$ is a dilation with scale factor 2, this shows that $AC = 2MP$.
Midpoints of Triangle Sides
Suppose $ABC$ is a triangle. Let $M$ be the midpoint of $\overline{AB}$ and $P$ the midpoint of $\overline{BC}$ as pictured below:
 Show that $\overleftrightarrow{MP}$ and $\overleftrightarrow{AC}$ are parallel.
 Show that $AC = 2MP$.
Comments
Log in to commentLynn says:
almost 3 yearsI am confused as to why this proof is an Illustration of GCO.C10. It is a valid proof and utilizes transformations but I believe that this proof should be reclassified as a GSRT proof
Math 1 students will be required to prove this theorem using the concepts of congruence and/or rigid motions. I have seen two different proofs that use these concepts.
One starts with drawing in a line parallel to side AB through point C and intersecting line MP.
The other one starts with rotating triangle MBP 180 degrees about point P
Michael Nakamaye says:
almost 3 yearsThe rationale for the alignment was that this result is explicitly stated in the GCO.10 standard but you are right that the method of proof is aligned with GSRT.4. It is a simpler version of https://www.illustrativemathematics.org/illustrations/1095 dealing with the special case of midpoints.