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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|.