Reading: BG, Section 8.3, Chapter 9, Sections 10.2 (lemma proof only), 10.3, 10.4
10-3 Exterior Angle Bisector Ratio Theorem
Prove: Given a triangle ABC, let m the exterior angle bisector line of angle ACB. If E is the point of intersection of m and line AB, then EA/EB = CA/CB.
Note: This proof should be self-contained. While you are welcome to use any ideas from that class activity and handout, you should not refer the reader to the handout for an understanding of your proof or your notation. A proof of the theorem for the interior angle bisector is in the reading in BG.
10-4 Squares on sides
(a) For a triangle ABC, construct squares on the sides of ABC with centers as shown in the figure. Then the composition X90Y90 is a rotation. Construct the center P of this rotation and tell what the angle of rotation is. Also, tell the measure of angle XPY.
(b) For a parallelogram ABCD, construct squares on the sides of ABCD (the squares should all be on the outside as with the triangle above). Prove that the centers of the squares on the sides are the vertices of a square. Hint: Use (a).
10-5 A point from a circle
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In this figure, O is the center of a circle of radius r. Line NS is the diameter perpendicular to line OA. Given point A, we construct point C in two steps: First, B is the intersection (distinct from N) of line NA with the circle. Second, C is the intersection of lines SB and OA. Question 1: If distance |OA| = d, find distance |OC| in terms if r and d and no other distances (if possible). Write the reasoning clearly. |
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Question 2. In the figure, A is outside the circle. Does your answer still hold if A is inside the circle or on the circle? Explain clearly.
10-6 Shorter Proof of Length-Minimizing Property of the Fermat Point
BG Section 10.4 is devoted to a proof of a key property of the Fermat point. There is a long proof on page 139 that triangle BQQ' is equilateral. The proof is so long because BG cannot call on the transformational tools that you have.
Task: State the Theorem of 10.4. Write a shorter, cleaner proof of the Theorem using a transformation to construct Q' and to give a quick proof that BQQ' is equilateral.
Note: The treatment in BG ends up with the Corollary, which we proved by other means in Lab 10.
10-7 Finding two centers of dilation of two circles
Review: Given two circles c and d, external to each other, we learned earlier how to find points E and I which are the intersections of the common external and internal tangents. Each of E and I is also the center of a dilation that carries c to d. The dilation with center E has a positive ratio and one with center I a negative ratio. The method was to construct two parallel diameters C1 C2 and D1 D2 (labeled paying attention to direction). Then E is the intersection of lines C1 D1 and C2 D2, and I is the intersection of lines C1 D2 and C2 D1.
Remark: This method produces centers of dilation E and I whether or not the circles are exterior to one another. But there are no common tangents through a center E or I if the center is interior to a circle.
Task: Construct the centers E and I for 6 examples of pairs of circles c and d in the figures on the attached page. In each case label the centers clearly E or I, to show which dilation has positive ratio and which has negative ratio. IT IS OK TO USE A DRAWING SQUARE FOR RIGHT ANGLES RATHER THAN A STRAIGHTEDGE AND COMPASS!
(a) Circles c and d are different sizes and exterior to one another.
(b) Circles c and d are different sizes and intersect one another.
(c) Circles d is interior to c, but the circles are not concentric.
(d) Circles c and d are the same size. (One of the points E or I may not exist in this
case.)
(e) Circles c and d are tangent internally.
(f) Circles c and d are tangent externally.
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Circles for Problem 10-7 Case (a) |
Circles for Problem 10-7 Case (b) |
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Circles for Problem 10-7 Case (c) |
Circles for Problem 10-7 Case (d) |
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Circles for Problem 10-7 Case (e) |
Circles for Problem 10-7 Case (f) |
