Lesson · Halving devices
Halving and doubling
A factor \(2\) in the statement - half a side, half the perimeter, twice a segment - is an instruction, not a decoration. It tells you to build a midpoint, or to double something. Three standard devices do almost all of this work.
The midline
Theorem
If \(M\) and \(N\) are the midpoints of the sides \(AB\) and \(AC\) of a triangle \(ABC\), then
Conversely, a segment through the midpoint of \(AB\) parallel to \(BC\) meets \(AC\) at its midpoint.
The trapezoid version follows at once: the segment joining the midpoints of the legs of a trapezoid with parallel sides \(a\) and \(b\) has length \(\tfrac{1}{2}(a+b)\).
The median to the hypotenuse
Theorem
In a right triangle with the right angle at \(C\), the midpoint \(M\) of the hypotenuse \(AB\) satisfies \(MC = MA = MB = \tfrac{1}{2}AB\). Conversely, if the median from \(C\) is half of \(AB\), the angle at \(C\) is right.
This is Thales' theorem wearing a different hat: \(M\) is the centre of the circle with diameter \(AB\), and \(C\) lies on that circle exactly when \(\angle ACB = 90^\circ\). The practical value is that it converts a right angle into two extra equal segments, and an isosceles triangle \(MAC\) whose base angles you can chase.
Doubling
Principle
Extending a segment beyond a midpoint by its own length creates a parallelogram. If \(M\) is the midpoint of \(BC\) and \(A'\) is the reflection of \(A\) in \(M\), then \(ABA'C\) is a parallelogram: \(A'C = AB\) and \(A'C \parallel AB\). The two sides that were far apart are now consecutive sides of one triangle.
Example
The median \(AM\) of a triangle satisfies \(AM < \tfrac{1}{2}(AB + AC)\). Double it: with \(A'\) the reflection of \(A\) in \(M\) we get \(A'C = AB\), and the triangle inequality in \(ACA'\) gives \(AA' < AC + CA' = AC + AB\). Since \(AA' = 2\,AM\), divide by \(2\).
Note what happened: the two sides \(AB\) and \(AC\), which share only the vertex \(A\), were brought into a single triangle by one reflection. That is the reason to double, every time.
Warm-up
Example
Let \(ABCD\) be a trapezoid with \(AB \parallel CD\), \(AB = a\), \(CD = b\) and \(a > b\). Let \(P\) and \(Q\) be the midpoints of the diagonals \(AC\) and \(BD\). Prove that \(PQ = \tfrac{1}{2}(a - b)\).
Where can a midline appear?
\(P\) and \(Q\) are midpoints of segments, but not of sides of one triangle - yet. Add the midpoint \(M\) of the leg \(AD\) and look at the triangles \(ACD\) and \(ABD\).
Two midlines on one line
In triangle \(ACD\), \(MP\) is a midline, so \(MP \parallel CD\) and \(MP = \tfrac{b}{2}\). In triangle \(ABD\), \(MQ\) is a midline, so \(MQ \parallel AB\) and \(MQ = \tfrac{a}{2}\). Both are parallel to \(AB\), so \(M\), \(P\), \(Q\) are collinear.
Finish by subtracting
\(P\) lies between \(M\) and \(Q\) because \(b < a\), so \(PQ = MQ - MP = \tfrac{a-b}{2}\).