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Geometry B Unit 7 Constructions Name Task #1 – Constructing the Center of a Circle (or an Arc) 1. Draw a chord. 2. Construct the extending the line through the circle. of the chord, This line goes through a diameter of the circle. State the theorem that justifies this. 3. Repeat this process one more time, creating a second diameter. The intersection of the two diameters is the center of the circle. Task #2 – Constructing the Circumscribed Circle of a Triangle B A C 1. Construct the perpendicular bisector of AC . Label this line n. By the Perpendicular Bisector Theorem, every point on line n is equidistant from A and C. 2. Construct the perpendicular bisector of BC . Label this line m. By the Perpendicular Bisector Theorem, every point on line m is equidistant from B and C. 3. The intersection of line n and line m is equidistant from A, B, and C. Label that point P. Therefore, because AP = BP = CP, P is the center of the circle that contains A, B, and C. The center of the circumscribed circle is the intersection of the of each of the sides of the triangle. This point is called the circumcenter. Task #3 – Constructing the Inscribed Circle of a Triangle B A C 1. Construct the angle bisector of A . Label this ray AX . By the Angle Bisector Theorem, every point on AX is equidistant from AB and AC . 2. Construct the angle bisector of C . Label this ray CY . By the Angle Bisector Theorem, every point on CY is equidistant from CB and AC . 3. The intersection of line n and line m is equidistant from AB , AC , and BC . Label that point P. The center of the inscribed circle is the intersection of the of each of the angles of the triangle. This point is called the incenter. Task #4 – Constructing a Square Inscribed in a Circle 1. Construct the center of the circle. Label it point O. (See Task #1) 2. Mark point A on the circle to be the first vertex of the square. 3. Using a straightedge, construct a diameter with an endpoint at A through the center of the circle, marking the other endpoint C. 4. Construct the perpendicular bisector of AC . Make sure the line intersects the circle twice. Label it line n. It should go through the center of the circle. Why? 5. Because the perpendicular bisector goes through the center of the circle, you have constructed another diameter. Label the endpoints B and D. 6. A, B, C, and D are the vertices of the square. With your straightedge, connect vertices to create the square. Prove that ABCD is a square. A, B, C, and D are points on the circle with center O, so AO BO CO DO because . Because diagonals AC and BD have the same midpoint, . Because ABCD is a parallelogram with AC BD , . Because ABCD is a parallelogram with AC BD , . Because ABCD is a and , ABCD is a square. Task #5 – Constructing an Equilateral Triangle Inscribed in a Circle 1. 2. 3. 4. 5. 6. 7. 8. 9. Construct the center of the circle. Label it point O. (See Task #1) Mark point A on the circle to be the first vertex of the triangle. Using a straightedge, construct the radius from A to O. Using your compass, construct a circle with the center on point A, with point O on the circle. Label one of the intersection points B. Using your compass, construct a circle with the center on point B, with point O on the circle. Label the new intersection point C. Continue this process until you have create circles A, B, C, D, E, and F . A, C, and E are vertices of the equilateral triangle. With your straightedge, connect vertices to create the triangle. Prove that ACE is an equilateral triangle. In the construction O A B C D E , so AB BC CD DE EF AF because . Because the chords AB BC CD DE EF AF , the central angles that intercept them are congruent, by Theorem. Therefore, AOB ____________ ____________ ____________ ____________ ____________ mAOB mBOC ______________ , mCOD mDOE ______________ , mEOF mFOA ______________ . As stated earlier, because these central angles are congruent, when you add them together, they are also congruent. Therefore, AOC ____________ ____________ By the Angle Addition Postulate, By the Congruent Circle Parts Theorem, if central angles AOC ____________ ____________ , their intercepted chords are congruent, AC _________ _________ . Because all three sides are congruent, ACE is equilateral. Task #6 – Constructing a Regular Hexagon Inscribed in a Circle 1. Follow steps #-18 for Task #5 2. A, B, C, D, E and F are vertices of the regular hexagon. With your straightedge, connect vertices to create the hexagon. Prove that ABCDEF is a regular hexagon. In the construction, O A B C D E , so AB BC CD DE EF AF and AO BO CO DO EO FO because . This makes 6 congruent equilateral triangles, AOB , ___________, ___________, ___________, ___________, and ___________ . This means that each angle in those triangles is . So, mABC mBCD mCDE mDEF mEFA mFAB _____________ Because m ABC m BCD m CDE m DEF m EFA m FAB , hexagon ABCDEF is equiangular. Because AB BC CD DE EF AF , hexagon ABCDEF is equilateral. Because hexagon ABCDEF is regular. and , it is