MATH325: Lecture 12

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More Constructions

Recall that last class we proved:

Lemma 3.3.2: We can construct the perpendicular bisector of any arbitrary line segment AB.

Can you remember the details of the construction? This class we'll start by proving something similar:

Lemma 3.3.1: If ∠BAC is a constructed angle, then we can bisect it.

Discussion of Proof: To bisect an angle, we must construct a line down the center of the angle -- in other words, the ray splits the angle into two congruent angles.

The construction is similar to that of 3.3.2. We first construct a point C' "opposite B" on the ray AC, then we use intersecting circles to construct a line AD that passes halfway between B and C'.

Once we've constructed the bisector, we must formally prove that ray AD is a bisector -- in other words, that ∠BAD is congruent to ∠DAC. We do this using congruent triangles.

Proof: We're given angle BAC. Construct the circle centered at A that passes through B; let C' be the point of intersection of that circle and ray AC. Then B and C' are equidistant from A.

Construct the circle centered at B that passes through C' and the circle centered at C' that passes through B. The circles intersect; let point D be one of the points of intersection; in particular, make sure D ≠ A.

Now me must prove that line AD bisects angle BAC. First, note that triangles ABD and AC'D are congruent by side-side-side. (Two pairs of sides are congruent because their lengths equal the lengths of radii of circles, and two are congruent because AD = AD.) Since triangles ABD and AC'D are congruent, it must be true that angles BAD and C'AD are congruent. Therefore, line AD bisects ∠BAC' (which equals ∠BAC).// Note that if you're familiar with the proof of Lemma 3.3.2, it's slightly easier to complete the proof by constructing the midpoint M of segment BC' and use side-side-side to prove that angle BAM is congruent to angle C'AM.

Last class I told you that we could construct distances by constructing two points whose separation equalled that distance, and that we could construct circles if we had a constructed or given point for the center and a construced or given distance for the radius.

However, looking closely at the rules given in the book we see that they insist that circles can only be constructed about an existing point and through a constructed or given point. The rules do not allow you to "set" your compass equal to a distance and then copy that distance to be the radius of a circle about some other point.

Does this mean that our notes from last class are wrong? Luckily, it doesn't.

Lemma 3.3.3: Suppose A, B and C are given or constructed points. Then we can construct a circle centered at A with radius |BC|.

Discussion of Proof: What we need to do is somehow "copy" distance |BC| to a segment ending in point A. Then our rules allow us to construct a circle centered at A whose radius is this segment. This would be easy if we were allowed to use Axiom 6, but Axiom 6 isn't one of the four rules we were given.

Since we can't use Axiom 6, we concoct a way to copy distance |BC| onto segment AB, then essentially reflect that distance across the midpoint of the segment. This lets us construct the desired circle.

The "reflection" process is different depending on the relative lengths of the segments, so the proof has two cases. Our proof will handle both cases, and so is slightly different from the one in the book.

Proof: Construct ray BA and the circle centered at B that passes through C. Let D be the point of intersection of the circle and the ray. Let M be the midpoint of segment BA.

Case 1: If |AB| > 2|BC|, then D is closer to B than to A. Construct a circle centered at M that passes through point D. This circle intersects segment BA at a second point D', as shown in Figure 3.3.

Since M is the midpoint of BA and since D and D' are equidistant from M, |D'A| = |DB|. But |DB| = |BC| by construction, so |D'A| = |BC|. Therefore, by constructing the circle centered at A through point D' we are constructing a circle with center A and radius |BC|. We have now completed the proof for the case in which |AB| > 2|BC|.

Case 2: If |AB| < 2|BC|, we must replace Figure 3.3 by a picture of two large overlapping circles about A and B and a smaller circle inscribed in their intersection. However, the proof proceeds in much the same way.

Construct the circle about M that passes through point D, and let D' be the other point of intersection of that circle with line AB. (Notice that D and D' do not lie on segment AB if |AB| < |BC|.) Whether D' lies on segment AB or not, |AD'| = |BD| because D and D' are equidistant from M. So once again the circle centered at A which passes through point D' is a circle with center A and radius |BC|.

Case 3: If |AB| = 2|BC| then construct the midpoint M of segment AB. In this case |AM| = |BC|.//

Extra credit: Prove that the following set of rules is equivalent to the rules given in the book: