Solutions for Session 4, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6| B7

Problem B1

 a. Answers will vary. Remember that the Line Tool has two arrows; do not use the Segment Tool. b. You need to select a line and a point. c. Use the Line Tool to draw a line anywhere on the sketch. Use the Point Tool to create a point that is not on the line. Select both objects (point and line -- remember to hold down the shift key!), and then choose "Parallel Line" from the Construct menu.

Problem B2

 a. Answers will vary. b. The points of intersection between the transversal and the pair of parallel lines will change, as will the measures of all of the angles. Also, the measures of linear pairs will add up to 180°. These are invariants for the diagram.

 Problem B3 Corresponding: ABF and EFH; CBF and GFH; DBC and BFG Alternate interior: EFB and CBF Vertical: EFH and BFG; ABF and DBC; BFE and GFH Linear pair: ABF and CBF; EFB and GFB; EFH and GFH; DBA and ABF; BFE and EFH; DBC and CBF; BFG and GFH

 Problem B4 Corresponding angles are congruent (have the same measure). Alternate interior angles are congruent. Vertical angles are congruent. Linear pairs are supplementary (angle measures add up to 180°).

Problem B5

 a. When you add m1 and m2 together, you define the same angle as the angle defined by a straight line, namely an angle of 180°. b. Same reasoning as in question (a). c. Using questions (a) and (b), we can write m1 + m2 = 180° m2 + m3 = 180° Subtracting the second equation from the first, we get m1 - m3 = 0 or m1 = m3 d. We can repeat the arguments from questions (a)-(c) with 2, 3, and 4, and conclude that 2 and 4 have the same measure.

 Problem B6 Let's consider 3 and 7 from the picture. We need to show that this pair of alternate interior angles has the same measure. Notice that 1 and 7 are corresponding angles, and therefore they have the same measure. In addition, 1 and 3 have the same measure because they are vertical angles (see Problem B5). And so it follows that 3 and 7 have the same measure because they both have the same measure as 1.

Problem B7

 a. The quadrilateral in the middle appears to be a parallelogram. b. Ideally, you want to test if opposite sides are parallel. Any of the following tests would suffice: Opposite sides are congruent; opposite angles are congruent; adjacent angles are supplementary (add up to 180°); the diagonals bisect each other. c. The two sides of the original quadrilateral opposite the vertex you choose and the angles between those sides remain the same as you move around the quadrilateral. Also, the side of the inscribed parallelogram opposite to the chosen vertex remains fixed. All other sides and angles change. The inscribed figure, however, remains a parallelogram. d. In part (c), you noticed that moving one vertex leaves two sides fixed. The reason for this is that they depend on the length of the diagonal. If we make the diagonals of the original quadrilateral perpendicular and congruent, then the inside quadrilateral will be a square. The resulting figure is a square inscribed in a kite or even a square, but that's not necessary.