### MATH325: Lecture 6

**Syllabus | Schedule | Grading Keys/Rubrics
**

#### Review of Homework

When a homework assignment has an instruction like "prove that f must be either a
reflection or a rotation", a good way to approach it is to look at the definitions
of reflection and rotation and prove that f must satisfy one definition or the other.
#### The Carpenter's Lemma (continued)

**Theorem 1.6.1 (Star Trek Lemma):** The measure of an inscribed
angle is half the angular measure of the arc it subtends.
Last, class, we proved this theorem for the case in which the circle
lies inside the inscribed angle. There are two cases we won't prove
it for: the center is to one side of the angle, and the center is
opposite the (obtuse) angle. In both cases it is still true, and the
proof is similar.

The third case is the one in which triangle BOC is degenerate -- in
other words, the inscribed angle BAC is a right angle. The fact
that when a right angle is inscribed in a circle, the hypotenuse is a
diameter of the circle is the key to the
"Carpenter's
Construction" of a circle -- an easy to describe practical
application of geometry. We will prove this special case of the Star
Trek lemma here.

**Carpenter's Lemma:** If ∠BAC is a right angle inscribed in a
circle, BC is a diameter of the circle.

**Discussion of Proof:** What we need to prove is that the circle's
center O lies on segment BC. The proof proceeds by contradiction -- assume that O
is on one side or the other of BC, assume that we've proved the Star
Trek Lemma for acute angles in that case, and use that to derive the contradiction that O
*must* lie on line BC..

**Proof:** Because triangle BAC is a right triangle, angles ABC and
ACB are acute angles. Assume BC is not a diameter of the circle -- i.e.
that O does not lie on line BC.

We're assuming we've proved the two acute angle cases of the Star Trek Lemma in which O does not lie on
AC, BC or AB; we conclude that:

m ∠AOC = 2 * m ∠ABC and

m ∠AOB = 2 * m ∠ACB.

We know that the sum of the angles of triangle ABC is 180° and
that angle BAC is a right angle, so:

m ∠ABC + m ∠ACB = 90°,

2 * (m ∠ABC + m ∠ACB) = 180°, and so

m ∠AOC + m ∠AOB = 180°.

Thus, AOC and AOB are supplementary angles and so point O must lie on
line BC. This contradicts our assumption that it does not, and so proves
that segment BC is a diameter of the circle.//

The carpenter's construction is an application of this lemma that
allows carpenters to accurately sketch the outlines of circular arches
or windows. Carpenters have large metal L-shaped rulers called
"carpenter's
squares". If a carpenter puts a nail at each end of the
diameter of a circle (say, at the top corners of a future doorway), he
or she can rest the ruler on the two nails and mark the location of
the right angle of the L.