Lesson 15-7: Lengths of Chords, Secants, and Tangents (Day 157)

Lesson 15-7 of the U of Chicago text is called "Lengths of Chords, Secants, and Tangents." In the modern Third Edition of the text, lengths of chords, secants, and tangents appear in Lesson 14-7. As for the new Lesson 14-6, this is a lesson that doesn't appear in the old text. It's all about three circles associated with a triangle -- the circumcircle, incircle, and nine-point circle. The first two of these circles are directly related to the concurrency theorems that appear in the Common Core Standards, so it's good that the authors include this lesson in the third edition.

Today is the final day in the Economics class. I never am able to play the Wall Street DVD, so instead I let them finish Boiler Room, the film they started on Friday. Unfortunately on this last day, the students try to break the cell phone and gum/food rules the most. This is always what I worry about in any classroom management situation -- the students follow the rules at first, then try to break them later on. This is something I must watch out for no matter what classroom I wind up in.

This is what I wrote last year about today's lesson:

Here are the two big theorems of this lesson:

Secant Length Theorem:
Suppose one secant intersects a circle at A and B, and a second secant intersects the circle at C and D. If the secants intersect at P, then AP * BP = CP * DP.

Given: Circle O, secants AB and CD intersect at P.
Prove: AP * BP = CP * DP.

There are two figures, depending on whether P is inside or outside the circle, but proofs are the same.

Proof:
Statements                    Reasons
1. Draw DA and BC.     1. Two points determine a line.
2. Angle BAD = BCD,  2. In a circle, inscribed angles intercepting
    Angle ADC = ABC       the same arc are congruent.
3. Triangle DPA ~ BPC 3. AA~ Theorem (steps 2 and 3)
4. AP / CP = DP / BP    4. Corresponding sides of similar
                                           figures are proportional.
5. AP * BP = CP * DP   5. Means-Extremes Property

This leads, of course, to the definition of power of a point.

Tangent Square Theorem:
The power of point P for Circle O is the square of the length of a segment tangent to Circle O from P.

Given: Point P outside Circle O and Line PX tangent to Circle O at T.
Prove: The power of point P for Circle O is PT^2.

Proof:
Statements                    Reasons
1. Draw Ray TO which 1. Two points determine a line.
intersects Circle O at B.
2. Let PB intersect         2. A line and a circle intersect in at most
Circle O at A and B.           two points.
3. PT perpendicular TB 3. Radius-Tangent Theorem and def, of semicircle
   and TAB in semicircle     
4. PTB right triangle     4. Definitions of right angle, right triangle,
with altitude TA,                and altitude
5. PT^2 = PA * PB        5. Right Triangle Altitude Theorem
6. The power of point P 6. Definition of power of a point
for Circle O is PT^2

By the way, we can now finally prove the Bonus Question from Lesson 15-4. I think I'll dispense with two-column proofs here and give a paragraph proof. We begin with a lemma:

Lemma:
Suppose two circles intersect in two points. Then for each point on their common secant line, the power of that point for first circle equals the power of that point for the second circle.

Given: Circles A and B intersect at C and D, Point P on secant CD
Prove: The power of point P for Circle A equals the power of point P for Circle B

Proof:
For both circles, the power of P is CP * DP, no matter whether P is inside or outside the circle. This common secant has a name -- the radical (or power) axis of the two circles. QED

Theorem:
Suppose each of three circles, with noncollinear centers, overlaps the other two. Then the three chords common to each pair of circles are concurrent.

Proof:
The proof of this is similar to that of the concurrency of perpendicular bisectors of a triangle (which I'll compare to this proof in parentheses). Let A, B, and C be the three circle centers. Every point on the radical axis of A and B has the same power for both circles. (Compare how every point on the perpendicular bisector of XY is equidistant from the points X and Y.) Every point on the radical axis of B and C has the same power for both circles. (Each point on the perpendicular bisector of YZ is equidistant from Y and Z.) So the point where these chords intersect has the same power for all three circles, and thus must lie on the radical axis of Circles A and C. (So the point where the perpendicular bisectors of XY and YZ must be equidistant from all three points, and so must lie on the perpendicular bisector of XZ.) The point where all three radical axes intersect is called the radical center (for perpendicular bisectors, it's called the circumcenter.) QED

Notice that if three centers are collinear, then the three radical axes are parallel (just as if three points are collinear, then the perpendicular bisectors are parallel).

I decided to make this the activity day for this week, since the Exploration section includes two questions rather than one. This is more like a puzzle, since the key to both questions isn't Geometry but arithmetic (or Algebra) and the properties of multiplication!