CHAPTER 5
Parallelograms, Trapezoids, Medians, and Midpoints

5.1 Trapezoids

A trapezoid is a quadrilateral having two, and only two, parallel sides. The bases of the trapezoid are its parallel sides; the legs are its nonparallel sides. The median of the trapezoid is the segment joining the midpoints of its legs.

Thus in trapezoid ABCD in Fig. 5-1, the bases are and , and the legs are and . If M and N are midpoints, then is the median of the trapezoid.

Fig. 5-1

An isosceles trapezoid is a trapezoid whose legs are congruent. Thus in isosceles trapezoid ABCD in Fig. 5-2 ≐ .

Fig. 5-2

The base angles of a trapezoid are the angles at the ends of its longer base: ∠A and ∠D are the base angles of isosceles trapezoid ABCD.

5.1A Trapezoid Principles

PRINCIPLE 1: The base angles of an isosceles trapezoid are congruent.

Thus in trapezoid ABCD of Fig. 5-3 if ≐ , then ∠A ≐ ∠D.

Fig. 5-3

PRINCIPLE 2: If the base angles of a trapezoid are congruent, the trapezoid is isosceles.

Thus in Fig. 5-3 if ∠A ≐ ∠D, then ≐ .

SOLVED PROBLEMS

5.1 Applying algebra to the trapezoid

In each trapezoid in Fig. 5-4, find x and y.

Fig. 5-4

Solutions

(a) Since || , (2x–5) + (x + 5) = 180; then 3x = 180 and x = 60.

Also, y + 70 = 180 or y = 110.

(b) Since ∠A ≐ ∠D, 5x = 3x + 20, so that 2x = 20 or x = 10.

since || , y + (3x + 20) = 180, so y + 50 = 180 or y = 130.

(c) Since || , 3x + 2x = 180 or x = 36.

Since ∠D ≐ ∠A, y = 2x or y = 72.

5.2 Proof of a trapezoid principle stated in words

Prove that the base angles of an isosceles trapezoid are congruent.

PROOF:

5.2 Parallelograms

A parallelogram is a quadrilateral whose opposite sides are parallel. The symbol for parallelogram is . Thus in ABCD in Fig. 5-5, || in || .

Fig. 5-5

If the opposite sides of a quadrilateral are parallel, then it is a parallelogram. (This is the converse of the above definition.) Thus if || and || , then ABCD is a .

5.2A Principles Involving Properties of Parallelograms

PRINCIPLE 1: The opposite sides of a parallelogram are parallel. (This is the definition.)

PRINCIPLE 2: A diagonal of a parallelogram divides it into two congruent triangles.

is a diagonal of ABCD in Fig. 5-6, so ΔI ≐ ΔII.

Fig. 5-6

PRINCIPLE 3: The opposite sides of a parallelogram are congruent.

Thus in ABCD in Fig. 5-5, ≐ and ≐ .

PRINCIPLE 4: The opposite angles of a parallelogram are congruent.

Thus in ABCD, ∠A ≐ ∠C and ∠B ≐ ∠D.

PRINCIPLE 5: The consecutive angles of a parallelogram are supplementary.

Thus in ABCD, ∠A is the supplement of both ∠B and ∠D.

PRINCIPLE 6: The diagonals of a parallelogram bisect each other.

Thus in ABCD in Fig. 5-7, ≐ and ≐ .

Fig. 5-7

5.2B Proving a Quadrilateral is a Parallelogram

PRINCIPLE 7: A quadrilateral is a parallelogram if its opposite sides are parallel.

Thus if || and || , then ABCD is a .

PRINCIPLE 8: A quadrilateral is a parallelogram if its opposite sides are congruent.

Thus if ≐ and ≐ in Fig. 5-8, then ABCD is a .

Fig. 5-8

PRINCIPLE 9: A quadrilateral is a parallelogram if two sides are congruent and parallel.

Thus if ≐ and || in Fig. 5-8, then ABCD is a

PRINCIPLE 10: A quadrilateral is a parallelogram if its opposite angles are congruent.

Thus if ∠A ≐ ∠C and ∠B ≐ ∠D in Fig. 5-8, then ABCD is a .

PRINCIPLE 11: A quadrilateral is a parallelogram if its diagonals bisect each other.

Thus if ≐ and ≐ in Fig. 5-9, then ABCD is a .

Fig. 5-9

SOLVED PROBLEMS

5.3 Applying properties of parallelograms

Assuming ABCD is a parallelogram, find x and y in each part of Fig. 5-10.

Fig. 5-10

Solutions

(a) By Principle 3, BC = AD = 3x and CD = AB = 2x; then 2(2x + 3x) = 40, so that 10x = 40 or x = 4.

By Principle 3, 2y–2 = 3x; then 2y–2 = 3(4), so 2y = 14 or y = 7.

(b) By Principle 6, x + 2y = 15 and x = 3y.

Substituting 3y for x in the first equation yields 3y + 2y = 15 or y = 3. Then x = 3y = 9.

(c) By Principle 4, 3x–20 = x + 40, so 2x = 60 for x = 30.

By Principle 5, y + (x + 40) = 180. Then y + (30 + 40) = 180 or y = 110.

5.4 Applying principle 7 to determine parallelograms

Name the parallelograms in each part of Fig. 5-11.

Fig. 5-11

Solutions

(a) ABCD, AECF; (b) ABFD, BCDE; (c) ABDC, CDFE, ABFE.

5.5 Applying principles 9, 10, and 11

State why ABCD is a parallelogram in each part of Fig. 5-12.

Fig. 5-12

Solutions

(a) Since supplements of congruent angles are congruent, opposite angles of ABCD are congruent. Thus by Principle 10, ABCD is a parallelogram.

(b) Since perpendiculars to the same line are parallel, || . Hence by Principle 9, ABCD is a parallelogram.

(c) By the addition axiom, ≐ . Hence by Principle 11, ABCD is a parallelogram.

5.6 Proving a parallelogram problem

PROOF:

5.3 Special Parallelograms: Rectangle, Rhombus, and Square

5.3A Definitions and Relationships among the Special Parallelograms

Rectangles, rhombuses, and squares belong to the set of parallelograms. Each of these may be defined as a parallelogram, as follows:

1. A rectangle is an equiangular parallelogram.

2. A rhombus is an equilateral parallelogram.

3. A square is an equilateral and equiangular parallelogram. Thus, a square is both a rectangle and a rhombus.

The relations among the special parallelograms can be pictured by using a circle to represent each set. Note the following in Fig. 5-13:

Fig. 5-13

1. Since every rectangle and every rhombus must be a parallelogram, the circle for the set of rectangles and the circle for the set of rhombuses must be inside the circle for the set of parallelograms.

2. Since every square is both a rectangle and a rhombus, the overlapping shaded section must represent the set of squares.

5.3B Principles Involving Properties of the Special Parallelograms

PRINCIPLE 1: A rectangle, rhombus, or square has all the properties of a parallelogram.

PRINCIPLE 2: Each angle of a rectangle is a right angle.

PRINCIPLE 3: The diagonals of a rectangle are congruent.

Thus in rectangle ABCD in Fig. 5-14, ≐ .

Fig. 5-14

PRINCIPLE 4: All sides of a rhombus are congruent.

PRINCIPLE 5: The diagonals of a rhombus are perpendicular bisectors of each other.

Thus in rhombus ABCD in Fig. 5-15, and are bisectors of each other.

Fig. 5-15

PRINCIPLE 6: The diagonals of a rhombus bisect the vertex angles.

Thus in rhombus ABCD, bisects ∠A and ∠C.

PRINCIPLE 7: The diagonals of a rhombus form four congruent triangles.

Thus in rhombus ABCD, ΔI ≐ ΔII ≐ ΔIII ≐ ΔIV.

PRINCIPLE 8: A square has all the properties of both the rhombus and the rectangle.

By definition, a square is both a rectangle and a rhombus.

5.3C Diagonal Properties of Parallelograms, Rectangles, Rhombuses, and Squares

Each check in the following table indicates a diagonal property of the figure.

5.3D Proving that a Parallelogram is a Rectangle, Rhombus, or Square

Proving that a Parallelogram is a Rectangle

The basic or minimum definition of a rectangle is this: A rectangle is a parallelogram having one right angle. Since the consecutive angles of a parallelogram are supplementary, if one angle is a right angle, the remaining angles must be right angles.

The converse of this basic definition provides a useful method of proving that a parallelogram is a rectangle, as follows:

PRINCIPLE 9: If a parallelogram has one right angle, then it is a rectangle.

Thus if ABCD in Fig. 5-16 is a and m∠A = 90°, then ABCD is a rectangle.

Fig. 5-16

PRINCIPLE 10: If a parallelogram has congruent diagonals, then it is a rectangle.

Thus if ABCD is a and ≐ , then ABCD is a rectangle.

Proving that a Parallelogram is a Rhombus

The basic or minimum definition of a rhombus is this: A rhombus is a parallelogram having two congruent adjacent sides.

The converse of this basic definition provides a useful method of proving that a parallelogram is a rhombus, as follows:

PRINCIPLE 11: If a parallelogram has congruent adjacent sides, then it is a rhombus.

Thus if ABCD in Fig. 5-17 is a and ≐ , then ABCD is a rhombus.

Fig. 5-17

Proving that a Parallelogram is a Square

PRINCIPLE 12: If a parallelogram has a right angle and two congruent adjacent sides, then it is a square.

This follows from the fact that a square is both a rectangle and a rhombus.

SOLVED PROBLEMS

5.7 Applying algebra to the rhombus

Assuming ABCD is a rhombus, find x and y in each part of Fig. 5-18.

Fig. 5-18

Solutions

(a) Since ≐ , 3x–7 = 20 or x = 9. Since ΔABD is equiangular it is equilateral, and so y = 20.

(b) Since ≐ , 5y + 6 = y + 20 or y = 3. Since ≐ , x = y + 20 or x = 23.

(c) Since bisects ∠A, 4x–5 = 2x + 15 = 10. Hence, 2x + 15 = 35 and m∠A = 2(35°)= 70°.

Since ∠B and ∠A are supplementary, y + 70 = 180 or y = 110.

5.8 Proving a special parallelogram problem

PROOF:

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

PROOF:

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 || l₂ || l₃ in Fig. 5-19 and segments a and b of transversal are congruent, then segments c and d of transversal 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 ΔABC in Fig. 5-20 if M is the midpoint of and || , then N is the midpoint of .

Fig. 5-20

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 ΔABC, if M and N are the midpoints of and then || and MN = AC.

PRINCIPLE 4: 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 is the median of trapezoid ABCD in Fig. 5-21 then || , || , and m = (b + b′).

Fig. 5-21

PRINCIPLE 5: The length of the median to the hypotenuse of a right triangle equals one-half the length of the hypotenuse.

Thus in rt. ΔABC in Fig. 5-22, if is the median to hypotenuse , then CM = AB; that is, ≐ ≐ .

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 , , and are medians of ΔABC 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

Solutions

(a) Since BE = ED and GC = CD, x = 8 and y = 7.

(b) Since BE = EA and CG = AG, 2x–7 = 45 and 3y + 4 = 67. Hence x = 26 and y = 21.

(c) Since AC = CE = EG and HF = FD = DB, x = 10 and y = 6.

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 and F is the midpoint of . Hence x = 17 and y = 36.

(b) By Principle 3, DE = AC and DF = BC. Hence x = 24 and y = 12.

(c) Since ABCD is a parallelogram, E is the midpoint of . Then by Principle 2, G is the midpoint of .

By Principle 3, x = (27) = 13 and y = (15) = 7.

5.12 Applying principle 4 to the median of a trapezoid

If is the median of trapezoid ABCD in Fig. 5-26,

Fig. 5-26

(a) Find m if b = 20 and b′ = 28.

(b) Find b′ if b = 30 and m = 26.

(c) Find b if b′ = 35 and m = 40.

Solutions

In each case, we apply the formula m = (b + b′). The results are

(a) m = (20 + 28) or m = 24

(b) 26 = (30 + b′) or b′ = 22

(c) 40 = (b + 35) or b = 45

5.13 Applying principles 5 and 6 to the medians of a triangle

Find x and y in each part of Fig. 5-27.

Fig. 5-27

Solutions

(a) Since AM = MB, is the median to hypotenuse . Hence by Principle 5, 3x = 20 and y = 20. Thus, x = 6 and y = 60.

(b) and are medians of ΔABC. Hence by Principle 6, x = (16) = 8 and y = 3(7) = 21.

(c) is the median to hypotenuse ; hence by Principle 5, CD = 15.

and are medians of ΔABC; hence by Principle 6, x = (15) = 5 and y = (15) = 10.

5.14 Proving a midpoint problem

PROOF:

SUPPLEMENTARY PROBLEMS

5.1. Find x and y in each part of Fig. 5-28

(5.1)

Fig. 5-28

5.2. Prove that if the base angles of a trapezoid are congruent, the trapezoid is isosceles.

(5.2)

5.3. Prove that (a) the diagonals of an isosceles trapezoid are congruent; (b) if the nonparallel sides and of an isosceles trapezoid are extended until they meet at E, triangle ADE thus formed is isosceles.

(5.2)

5.4. Name the parallelograms in each part of Fig. 5-29.

(5.4)

Fig. 5-29

5.5. State why ABCD in each part of Fig. 5-30 is a parallelogram.

(5.5)

Fig. 5-30

5.6. Assuming ABCD in Fig. 5-31 is a parallelogram, find x and y if

(5.3)

Fig. 5-31

(a) AD = 5x, AB = 2x, CD = y, perimeter = 84

(b) AB = 2x, BC = 3y + 8, CD = 7x–25, AD = 5y–10

(c) m∠A = 4y–60, m∠C = 2y, m∠D = x

(d) m∠A = 3x, m∠B = 10x–15, m∠C = y

5.7. Assuming ABCD in Fig. 5-32 is a parallelogram, find x and y if

(5.3)

Fig. 5-32

(a) AE = x + y, EC = 20, BE = x–y, ED = 8

(b) AE = x, EC = 4y, BE = x–2y, ED = 9

(c) AE = 3x–4, EC = x + 12, BE = 2y–7, ED = x–y

(d) AE = 2x + y, AC = 30, BE = x + y, BD = 24

5.8. Provide the proofs requested in Fig. 5-33.

(5.6)

Fig. 5-33

5.9. Prove each of the following:

(a) The opposite sides of a parallelogram are congruent (Principle 3).

(b) If the opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram (Principle 8).

(c) If two sides of a quadrilateral are congruent and parallel, the quadrilateral is a parallelogram (Principle 9).

(d) The diagonals of a parallelogram bisect each other (Principle 6).

(e) If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram (Principle 11).

5.10. Assuming ABCD in Fig. 5-34 is a rhombus, find x and y if

(5.7)

Fig. 5-34

(a) BC = 35, CD = 8x–5, BD = 5y, m∠C = 60°

(b) AB = 43, AD = 4x + 3, BD = y + 8, m∠B = 120°

(c) AB = 7x, AD = 3x + 10, BC = y

(d) AB = x + y, AD = 2x–y, BC = 12

(e) m∠B = 130°, m∠1 = 3x–10, m∠A = 2y

(f) m∠1 = 8x–29,m∠2 = 5x + 4, m∠D–y

5.11. Provide the proofs requested in Fig. 5-35.

(5.8)

Fig. 5-35

5.12. Prove each of the following:

(5.9)

(a) If the diagonals of a parallelogram are congruent, the parallelogram is a rectangle.

(b) If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.

(c) If a diagonal of a parallelogram bisects a vertex angle, then the parallelogram is a rhombus.

(d) The diagonals of a rhombus divide it into four congruent triangles.

(e) The diagonals of a rectangle are congruent.

5.13. Find x and y in each part of Fig. 5-36

(5.10)

Fig. 5-36

5.14. Find x and y in each part of Fig. 5-37.

(5.11)

Fig. 5-37

5.15. If is the median of trapezoid ABCD in Fig. 5-38

(5.12)

Fig. 5-38

(a) Find m if b = 23 and b′= 15.

(b) Find b′ if b = 46 and m = 41.

(c) Find b if b′ = 51 and m = 62.

5.16. Find x and y in each part of Fig. 5-39.

(5.11 and 5.12)

Fig. 5-39

5.17. In a right triangle

(5.13)

(a) Find the length of the median to a hypotenuse whose length is 45.

(b) Find the length of the hypotenuse if the length of its median is 35.

5.18. If the medians of ΔABC meet in D

(5.13)

(a) Find the length of the median whose shorter segment is 7.

(b) Find the length of the median whose longer segment is 20.

(c) Find the length of the shorter segment of the median of length 42.

(d) Find the length of the longer segment of the median of length 39.

5.19. Prove each of the following:

(5.14)

(a) If the midpoints of the sides of a rhombus are joined in order, the quadrilateral formed is a rectangle.

(b) If the midpoints of the sides of a square are joined in order, the quadrilateral formed is a square.

(c) In ΔABC, let M, P, and Q be the midpoints of , , and , respectively. Prove that QMPC is a parallelogram.

(d) In right ΔABC, m∠C = 90°. If Q, M, and P are the midpoints of , , and , respectively, prove that QMPC is a rectangle.