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1 Suggested problems - solutions Circles (terminology) Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, In particular, see section 3.8, pp The problems are all from section 3.7. Problems: 5,6,8,9 #5: A chord half as far from the center of the circle as another chord has twice the length can be disprove. Remeber that it s easier to disprove a suspected false statement than it is to prove a suspected true one - all it takes is one counterexample, and the counterexample can be very specific. Verify that the numbers in the circle below all hold in Euclidean geometry (use Pythagorean theorem for distance as needed). The circle shown above has radius 5. Chord EB is located 2 units from the center of the circle, and has length EB = Chord CD is located 4 units from the center of the circle, and has length EB = 2(3) = 6. The chord which is half the distance (EB) isnottwicethe length!

2 #6: For this one, since you re asked to prove your answer, you need to figure out what the general principle is: if you have two chords of equal length which do not intersect inside the circle, and they are extended until they meet in a point outside the circle, the distance from the point of intersection of the chords to their respective points on the circle is also equal; i.e., in the picture below, if two chords are congruent with length y, show that x = z. So, take as the given: a circle centered at O with points A, B, C, D on the circle. Chords AB and CD can be extended to intersect at a point P in the exterior of the circle; i.e. P A B and P C D. AB = CD. The next move is to construct perpendiculars from AB to point O, andfromcd to point O (this is the unnamed theorem 4 on p a good name for it would be the Existence of a perpendicular through a point theorem):

3 Label the points of intersection L and M. By the definition of distance from a line (segment) to a point, the lengths OL and OM represent the distances from chord AB to O, andchordcd to O, respectively. By circle property five, two chords with the same length are equidistant from the center, so OL = OM: Now, consider the triangles formed by drawing in segement OP. Triangles POL and POM are right triangles, with hypoteneuse OP, since PLO and PMO are known to be right angles from the construction of the perpendiculars. The next move is to show that these triangles are congruent:

4 Since OP = OP (reflexivity), POL = POM by the HL (Hypoteneuse-Leg) Theorem for right triangles. Now, claim some parts: PL = PM by CPCF. The rest of it is just setting up and justifying the obvious; AL and CM should be the same length (but we have to justify that!), and then a bunch of algebra should give x = z.

5 Circle property three says that a line passing through the center of the circle and perpendicular to the chord bisects the chord. So OL bisects AB, and since AB = y, AL = 2 1 y. This also tells us the obvious A L B. Similarly, OM bisects CD, and since CD = y, CM = 1 2 y,andc M D Since P A B and A L B, we have P A L (can you find that theorem?), and PL = PA+ AL, orpl = z y. Since P C D and C M D, wehavep C M, andpm = PC+ CM,orPL = x y. Since PL = PM, by substitution z y = x y. And, finally x = z Finally, finally, the answer to the question in the book (which had some actual numbers in it!) is that for that particular example, x =6. Another point of interest is that this could be proven in a totally different way using parallel lines and similar triangles, if we were in Euclidean geometry (try it!)

6 Taking this idea a bit further, what are the results if two chords of the same length intersect in the interior ofthecircle-givenab = CD, AB and CD intersect in point P,withA P B and C P D, can you prove that x = z and w = y? This would make a very nice question for the takehome part of the exam, and could be something to start thinking about! (Hint. Hint.) #8: Circles K and T intersect, forming a common chord MN (text typo in my text, btw, the chord is a segment, not a ray!). Prove that the line of centers KT is the perpendicular bisector of MN. This is a quick proof if you recall the Perpendicular Bisector Theorem on p 144. Because KM and KN are both radii of the circle centered at K, KM = KN and therefore point K is equidistant from M and N. SoK lies on the perpendciular bisector of MN. Because TM and TN are both radii of the circle centered at T, TM = TN and therefore point T is equidistant from M and N. SoT lies on the perpendciular bisector of MN. K and T are both points on the perpendicular bisector of MN. Since two points determine a unique line (incidence axiom 1), the line KT must be the perpendicular bisector of MN

7 #8: Using the previous figure, prove KMT = KNT. Another quick one, and does not rely on the previous result (i.e. you don t need the info about the perpendicular bisector). You can go straight for congruent triangles. Since KM and KN are both radii of the circle centered at K, KM = KN. Since TM and TN are both radii of the circle centered at T, TM = TN. KT = KT (reflexivity). KMT = KNT by SSS theorem. So KMT = KNT by CPCF.

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