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(a) Since AB > AD, 3x (b) Since BC > AB, 5y 7 6 20 or x y 9. Since nABD is equiangular it is equilateral, and so y 31. 2 Since CD > BC, x y 20 or x 231. 2 2(35 ) 70 . 20.
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(c) Since AC bisects /A, 4x 5 2x 15 or x Since /B and /A are supplementary, y 70
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10. Hence, 2x 15 180 or y 110.
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5.8 Proving a special parallelogram problem
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Given: Rectangle ABCD E is midpoint of BC. To Prove: AE > ED Plan: Prove nAEB > nDEC
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1. 2. 3. 4. 5. 6. 7.
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Statements ABCD is a rectangle. E is midpoint of BC. BE > EC /B > /C AB > CD nAEB > nDEC AE > ED
Reasons 1. 2. 3. 4. 5. 6. 7. Given Given A midpoint divides a line into two congruent parts. A rectangle is equiangular. Opposite sides of a ~ are congruent. SAS s Corresponding parts of congruent n are congruent.
CHAPTER 5 Parallelograms, Trapezoids, Medians, and Midpoints
5.9 Proving a special parallelogram problem stated in words Prove that a diagonal of a rhombus bisects each vertex angle through which it passes.
Solution
Rhombus ABCD AC is a diagonal. To Prove: AC bisects /A and /C. Plan: Prove (1) /1 and /2 are congruent to /3. (2) /3 and /4 are congruent to /1.
PROOF:
Given:
1. 2. 3. 4. 5. 6. 7.
Statements ABCD is a rhombus. AB > BC /1 > /3 BC i AD, AB i CD /2 > /3, /1 > /4 /1 > /2, /3 > /4 AC bisects /A and /C.
Reasons 1. 2. 3. 4. 5. 6. 7. Given A rhombus is equilateral. In a n, angles opposite congruent sides are congruent. Opposite sides of a ~ are i. Alternate interior of i lines are congruent. Things congruent to the same thing are congruent to each other. To divide into two congruent parts is to bisect.
5.4 Three or More Parallels; Medians and Midpoints
5.4A Three or More Parallels
PRINCIPLE 1:
If three or more parallels cut off congruent segments on one transversal, then they cut off congruent segments on any other transversal.
Thus if l1 i l2 i l3 in Fig. 5-19 and segments a and b of transversal AB are congruent, then segments c and d of transversal CD are congruent.
Fig. 5-19
5.4B Midpoint and Median Principles of Triangles and Trapezoids
PRINCIPLE 2:
If a line is drawn from the midpoint of one side of a triangle and parallel to a second side, then it passes through the midpoint of the third side.
Thus in nABC in Fig. 5-20 if M is the midpoint of AB and MN i AC, then N is the midpoint of BC.
PRINCIPLE 3:
If a line joins the midpoints of two sides of a triangle, then it is parallel to the third side and its length is one-half the length of the third side.
Thus in nABC, if M and N are the midpoints of AB and BC, then MN i AC and MN
CHAPTER 5 Parallelograms, Trapezoids, Medians, and Midpoints
Fig. 5-20
PRINCIPLE 4:
Fig. 5-21
The median of a trapezoid is parallel to its bases, and its length is equal to one-half of the sum of their lengths.
Thus if MN is the median of trapezoid ABCD in Fig. 5-21 then MN i AD, MN i BC, and m
PRINCIPLE 5:
b ).
The length of the median to the hypotenuse of a right triangle equals one-half the length of the hypotenuse.
1 2AB;
Thus in rt. nABC in Fig. 5-22, if CM is the median to hypotenuse AB, then CM
that is, CM > AM > MB.
Fig. 5-22
PRINCIPLE 6:
The medians of a triangle meet in a point which is two-thirds of the distance from any vertex to the midpoint of the opposite side.
Thus if AN, BP, and CM are medians of nABC in Fig. 5-23, then they meet in a point G which is two-thirds of the distance from A to N, B to P, and C to M.
Fig. 5-23
SOLVED PROBLEMS
5.10 Applying principle 1 to three or more parallels Find x and y in each part of Fig. 5-24.
Fig. 5-24
CHAPTER 5 Parallelograms, Trapezoids, Medians, and Midpoints
Solutions
(a) Since BE (b) Since BE (c) Since AC ED and GC EA and CG CE
1 2 CD,
8 and y 7
72. 4 67. Hence x 6. 26 and y 21.
AG, 2x FD
45 and 3y DB, x
EG and HF
10 and y
5.11 Applying principles 2 and 3 Find x and y in each part of Fig. 5-25.
Fig. 5-25
Solutions
(a) By Principle 2, E is the midpoint of BC and F is the midpoint of AC. Hence x (b) By Principle 3, DE
1 2 AC
17 and y
and DF
1 2 BC.
Hence x 71 . 2
24 and y
121. 2
(c) Since ABCD is a parallelogram, E is the midpoint of AC. Then by Principle 2, G is the midpoint of CD. By Principle 3, x
1 2 (27)
131 and y 2
1 2 (15)
5.12 Applying principle 4 to the median of a trapezoid If MP is the median of trapezoid ABCD in Fig. 5-26, (a) Find m if b (b) Find b if b (c) Find b if b 20 and b 30 and m 35 and m 28. 26. 40.
Fig. 5-26
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