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Midpoints of Triangle Sides


Alignments to Content Standards: G-CO.C.10

Task

Suppose $ABC$ is a triangle. Let $M$ be the midpoint of $\overline{AB}$ and $P$ the midpoint of $\overline{BC}$ as pictured below:

Mpoint_b657db1c553cace8d748bc10afd31f65

  1. Show that $\overleftrightarrow{MP}$ and $\overleftrightarrow{AC}$ are parallel.
  2. Show that $|AC| = 2|MP|$.

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 G-SRT.1a and is an axiom in the transformational approach to geometry. Also important in this task is G-SRT.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 (non-zero) factor, and take lines not passing through the center of dilation to parallel lines.

Solution

  1. 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 $\overleftrightarrow{AC}$.

  2. Since $D\left(\overline{MP}\right) = \overline{AC}$ and $D$ is a dilation with scale factor 2, this shows that $|AC| = 2|MP|$.

Lynn says:

over 3 years

I am confused as to why this proof is an Illustration of G-CO.C10. It is a valid proof and utilizes transformations but I believe that this proof should be reclassified as a G-SRT 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:

over 3 years

The rationale for the alignment was that this result is explicitly stated in the G-CO.10 standard but you are right that the method of proof is aligned with G-SRT.4. It is a simpler version of https://www.illustrativemathematics.org/illustrations/1095 dealing with the special case of midpoints.