## Posts Tagged ‘Midline’

### Monday Math 156

July 28, 2014

For a trapezoid, the median (also known as the midline or mid-segment) is the segment connecting the midpoints of the legs of the trapezoid. This post presents a proof that:

1. the median of a trapezoid is parallel to the bases;
2. the length of the median is half the sum of the lengths of the bases;
3. the midpoints of the diagonals of a trapezoid also lie on its midline.

### Monday Math 155

July 21, 2014

The triangle midline theorem, also called the midsegment theorem, states that the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. This may be trivially proven via triangle similarity (SAS similarity condition) and the corresponding angles postulate. More interesting, however, is to prove it using triangle congruence.

Let D and E be the midpoints of sides AB and AC, respectively, of ∆ABC. Let us extend segment DE past E to point F such that DE=EF, and let us draw CF.

Since DE=EF, AE=EC, and vertical angles ∠AED and ∠CEF are congruent, we see by the SAS condition that ∆ADE≅∆CFE. Thus, CF=AD=BD. Also, ∠FCE≅∠DAE; but since these are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AD}$ and $\stackrel{\longleftrightarrow}{CF}$ cut by transversal $\stackrel{\longleftrightarrow}{AC}$, we see that $\stackrel{\longleftrightarrow}{AD}\parallel\stackrel{\longleftrightarrow}{CF}$. But then the quadilateral BCFD has a pair of opposite sides, BD and CF, which are of equal length and parallel, so it is therefore a parallelogram, and so $\stackrel{\longleftrightarrow}{DE}\parallel\stackrel{\longleftrightarrow}{BC}$. And since opposite sides of a parallelogram have equal length, DF=BC, and so DEDFBC.

We can also prove a similar theorem: that the line through the midpoint of one side of a triangle parallel to a second side of the triangle bisects the third side, and that the segment of that line inside the triangle is one-half as long as the parallel side. As with the midpoint theorem, it is trivial to prove with triangle similarity (this time the AA condition) and the corresponding angles postulate. So, instead, we use a similar construction as above.

Let D be the midpoint of side AB of ∆ABC, and let E be the point where the line through D parallel to BC intersects AC. Construct the line through C parallel to AB, and let F be the point where it intersects $\stackrel{\longleftrightarrow}{DE}$

Then BCFD is a parallelogram, and since opposite sides of a parallelogram have the same length, BD=CF and BC=DF. And so CF=BD=AD. And since ∠ADE and ∠CFE are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AD}$ and $\stackrel{\longleftrightarrow}{CF}$ cut by transversal $\stackrel{\longleftrightarrow}{DF}$, they are congruent. Similarly, ∠DAE≅∠CFE, and so, by the ASA condition, ∆ADE≅∆CFE. Thus, AE=EC, and E is thus the midpoint of AC. We see also that DE=EF, and since DF=BC, thus DEDFBC.

Using both of these theorems together, we can prove a third: that a midline (mid-segment) of a triangle and the triangle median that intersects it bisect each other.

Let MA, MB and MC be the midpoints of sides BC, AC and AB, respectively, of ∆ABC. Thus, $\overline{M_{B}M_{C}}$ is a midline of ∆ABC, and $\overline{AM_{A}}$ a median. Let P be the point where they intersect.

By the midline theorem, $\stackrel{\longleftrightarrow}{M_{B}M_{C}}\parallel\stackrel{\longleftrightarrow}{BC}$ and MBMCBC.

This means, then, that $\stackrel{\longleftrightarrow}{M_{C}P}\parallel\stackrel{\longleftrightarrow}{BM_{A}}$, and so by our second theorem above with regards to ∆ABMA, we see that P must be the midpoint of $\overline{AM_{A}}$, and MCPBMA.

Similarly, our second theorem applied to triangle AMAC establishes that MBPCMA. But BMA=CMA, and so MBP=MCP, and P is the midpoint of $\overline{M_{B}M_{C}}$ as well.