Note: In class we talked about Problem 22 and mentioned that these 22 problems had the instruction of working them without use of the parallel postulate, as was the case for all of Section 1.2.
PROBLEM SET 1.2
1. Symbolic representations of two congruent triangles.
2. Congruence condition for two quadrilaterals.
This is to be a general quadrilateral. Since are congruence conditions are for triangles, it seems some way to use SAS, SSS, or ASA must be found.
3. State and Prove a theorem analogous to Theorem 1.7 for non-convex kites.
See the proof discussion in Sec 1.2 for Thm1.7. If the line of the diagonal, rather than the segment, is considered, the proof deals with both the convex and non-convex kites.
4. Proof of Theorem 1.8
5. Proof of Theorem 1.9
6. Construction and Proof: The Perpendicular bisectors of the sides of a triangle are concurrent.
GSP File for Perpendicular Bisectors of a triangle
GSP File for Circumcircle of a triangle
See also Link to Jim's GSP Library
7. (NOT in PDF version) Draw a line l and a point P not on l. Construct a perpendicular to l through P. Describe the construction steps and prove the steps are valid (i.e., that the line you construct is perpendicular to l.
NOTE: This is asking for the construction with straightedge and compass. The construction tool on GSP already has the steps built into it. So the task here is to show the steps at a more fundamental level.
8. A circle that passes through each vertex of a triangle is called the circumscribed circle or circumcircle. A circle that is tangent to each side of a triangle is called an inscribed circle or an incircle.
a. Assume that a tangent to a circle is perpendicular to the radius at the point of contact. Explain how to find the circumscribing circle and the inscribed circle for a given triangle.
Solution and discussion by Brenda King
b. Draw a triangle in which (i) all the angles are acute; (ii) one angle is obtuse; and (iii) one angle is 90 degrees. In each case construct the circumscribing and inscribed circles.
c. What seems to be true about the centers of circumscribed circles.
9.
a. Prove that the angle bisectors of a rhombus are concurrent (intersect in a single point). Then prove that the point where the angle bisectors meet is equidistant from all sides of the rhombus.
b. Construct the rhombus and the circle inscribed in the rhombus. Clearly identify the radius of the circle.
10.
a. Prove the angle bisectors of a complex kite are concurrent.
b. Construct any convex kite and the circle inscribed in the kite. Clearly indicate the radius of the circle. Describe the construction and prove that it is valid.
11. (Problem 9 on PDF) Use the Exterior Angle Theorem to prove that in any triangle ABC,
m(Angle A) + m(Angle B) + m(Angle C) < 270 degrees.
Hint for proof: At most one of the angles can be more than 180 degrees. If one is more than 180 degrees, let that be Angle C. By Theorem 1.10 the measure each remote interior angle is less that the exterior angle. Consider the exterior angle at A and CAD be the exterior angle. Then angle C is 180-m(angle A). By Theorem 1.10,
m(angle B) < 180 - m(angle A)
m(angle C) < 180 - m(angle A)
m(angle A) + m(angle A + m(angle B) + m(angle C) < 360 degrees
But m(angle A) < 90 degrees, so
m(angle A) + m(angle B) + m(angle C) < 270 degrees
12. (Problem 10 in PDF) Prove that in any quadrilateral, the sum of the lengths of any three sides is greater than the length of the fourth side.
Construct quadrilateral ABCD, construct diagonal AC, and label as above.
By triangle inequality, j + k > n, n + m > l
Therefore j + k + m > l
In a similar manner, using diagonal BD when needed, the result can be shown for each of the sides.
13. Prove Theorem 1.14 (which is the converse of theorem 1.13), which states that if alpha > beta, then a > b, as follows.
14
a. Construct a scalene (not isosceles) triangle and another triangle congruent to in using the three sides of your triangle. Describe your construction and explain why it is valid.
b. Draw and angle and a ray that does not intersect the angle. Then use your construction in part a above construct an angle congruent to the original angle that has the ray as one of its sides. Briefly explain the idea behind the construction.
c. Repeat part (b), but this time use an isosceles triangle to "duplicate" the angle.
15. Construct a scalene triangle and a triangle congruent to it using only two sides and included angle of the original triangle.
16.
17. Given three segments of length a, b, and c, what conditions must a, b, and c satisfy to form the sides of some triangle.
Triangle inequality: a + b > c, b + c > a, a + c > b
18. Prove that the sum of the distances from any point in the interior of a triangle to the three vertices is greater than half the perimeter (the sum of the lengths of the sides) of the triangle.
Proof: By triangle inequality n + o > j, n + m > l, and m + o > k. Therefore 2(m + n + o) > a + b + c
19. Prove that the perimeter of a quadrilateral is greater than the sum of the lengths of the diagonals.
Proof: Use triangle inequality. j + k > n, l + m > n, k + l > o, m + j > o. Therefore 2(j + k + l + m) > 2(n + o)
Another solution:
Proof from Clay Kitchings
21. Prove that if a, b, and c are the length of the sides of a triangle, then there exists a triangle with sides of length
,
, and
.
Proof: Following the suggestion of the author, we write, for x > 0, y > 0
The sides of the triangle a, b, c are positive values. By the Triangle inequality a + b > c, b + c > a, and c + a > b. By the above, then
>
>
>
22. Angle AOB has vertex O, which is not on the paper. Construct the bisector of angle AOB (without extending the sides on additional paper). Describe your construction and prove that it produces the required angle bisector.
Problem 1.2.22 construction suggested by Chelsea Henderson.
This IS a valid approach IF we have assumed the Parallel Postulate and make that a part of our geometry. We will do that in Section 1.3. For this problem set, however, we have the restriction to use only the axioms and theorems up through Section 1.2 prior to the assumption of the parallel postulate.
A Solution from Brenda King
