Efj and with same direction

FUNDAMENTALS OF COLLEGE GEOMETRY

196

^POLYGONS

-

PARALLELOGRAMS

197

12. Prove that line segments drawn from A and B of LABC to the op- posite sides cannot bisect each other.

(Hint: Use indirect method by assuming AS and RB bisect each other; then ABSR is aD, ete.)

c

~

A B

L

^{G //}

^H/

^//

^K

B L-_-

A ^/

/ / / /

EL _D

Ex. 12.

13. Prove that a quadrilateral is a rhombus if the diagonals bisect each other and are perpendicular to each other.

14. Prove that if from the point where the bisector of an angle of a triangle meets the opposite side parallels to the other sides are drawn a rhombus is formed.

6.17. Direction of rays. Two rays have the same direction if and only if either they are parallel and are contained in the same closed half-plane determined by the line through their endpoints or if one ray is a subset of the other (Fig. : 6.8).

Theorem 6.8.

Given: LABC and LDEF with

'l3t

II

EP

and with same direction;

198

STATEMENTS REASONS

Proof

The proof is left as an exercise for the student.

Theorem 6.10

6.20. The segment joining the midpoints of two sides of a triangle is parallel to the third side, and its measure is one half the measure of the third side.

Given: 6ABC with the M the midpoint of AC and N the midpoint of BC.

Conclusion: MN

II

AB, mMN = tmAB.

Proof

ST A TEMENTS

c

A

Theorem 6.10.

REASONS

I I

1. Point plotting postulate.

1. On the ray opposite NM con-

struct ND such that ND ~ MN.

2. Draw BD.

3. N is the midpoint of BC.

4. NB ~

Ne.

5. LDNB

^~ LMNC.

6. 6DNB ~

6MNC.

7. BD

^~

CM.

8. M is the midpoint of AC.

9. CM ~ AM.

10. BD ~ AM.

11. LDBN ~ LMCN.

12. BD

_II

AC.

13. ABDM is a D.

14. MN

II

AB.

15. MD

^~

AB.

16. mMD = mAB.

17. mMD=mMN+mND.

2. Postulate 2.

3. Given.

4. Definition of midpoint.

5. Vertical angles are ~.

6. S.A.S.

7. Corresponding sides of ~

& are -.

8. Given.

9. Reason 4.

^N

10. Theorem 4.3.

11. Reason 7.

12. Theorem 5.11.

13. Theorem 6.5.

14. Definition of aD.

,

15. The opposite sides of a CJ are 16. Definition

ments.

17. Postulate perty.

18. mAB = mMN +mMN. 18. Theorem 3.5 and substitution property.

19. Symmetric property of equality.

20. Division property of equality.

19. mA1N+ mMN = mAB.

20. mMN = tmAB.

Theorem 6.11

6.21. A line that bisects one side of a tri- angle and is parallel to a second side bisects the third side.

c

A

L/)~

R B

Given: MNbisects AC,

MN/iAB.

Conclusion: M1Vbisects Be.

Proof

Theorem 6.11.

ST ATEMENTS REASONS

The proof is left as an exercise for the student.

Theorem 6.12

6.22. The midpoint of the hypotenuse of a right triangle is equidistant from its vertices.

c

Given: Right 6ABC with LABC a right L;

M is the midpoint of Ae.

Conclusion: AM ~ BM ~ CM.

Proof

A

STATEMENTS REASONS Theorem 6.12.

The proof is left as an exercise for the student.

Then prove 6BMN

^~

6CMN.) (Hint: Draw MN

II

AB.

Exercises

1. Given: 6ABC with R, S, T midpoints of AC, BC, and AB respectively;

200 FUNDAMENTALS OF COLLEGE GEOMETRY

mAC

= 6 inches;

mBC = 8 inches;

mAB = 12 inches.

the value of mST+mRT.

Find: mRS+

2. Given: LC is a right L; mLA = 60;

M is the midpoint of AB, mAC = 8 inches.

the value of mAB.

Find:

3. Given: AB

II

DC; M is the mid- point of AD; N is the mid- point of BC.

Prove: 1'\1N

II

4B;lUN

^Ii

oc;

mMN = t(mAB + mCD) . (Hint: Draw i5N until it meets AB at, say, P.) 4. In the figure for Ex. 3, find the

length of AB if mDC = 8 feet, mMN = 11 feet.

5. Given: Quadrilateral ABCD with Q, R, S, P, the midpoints of AB, BC, CD, and DA respectively.

PQRSisa D.

(Hint: You will need to draw the diagonals of ABCD.)

Prove:

---

A

c

/o~:

A T B.1

EX.i.

B

A Ex. 2.

D

Exs. 3, 4.

D

---

6. Given:

Quadrilateral KLMN with P, Q, R, S the midpoints of KL, LM,MN,andNK.

PRand QS bisect each other.

Prove:

7. Given: ABCD is aD; K, L, M, N are midpoints of OC, OD, OA, OB.

Prove: KLMN is a D.

A

8. Given: MN is the median of trapezoid ABCD; AC and BD are dia- gonals.

Prove: MN bisects the diagonals.

9. Given: BN and AM are medians of LABC; L is the midpoint of OA; K is the mid point of OB.

Prove: KMNL is aD.

10. Prove that the line joining the mid- points of two opposite sides of a parallelogram bisects the diagonal of the parallelogram.

11. Prove that the lines joining the mid- Points of the sides of a rectangle form a rhombus.

POLYGONS-PARALLELOGRAMS 201 N

M

L Ex. 6.

c

Ex. 7.

8 M B

Ex.8.

c

A B

Ex.9.

I

Parallelogram

I

Rectangle

I

Square /Rhombus!

Both pairs of opposite sides are parallel.

TRliE-FAI.SE STATEMENTS

1. The diagonals of a parallelogram

bisect each other:

2. A quadrilateral that has two and only two parallel sIdes IS a rhombus.. 3. The bases of a trapezoid are parallel to each other.

4. An equilateral parallelogram is a square.

5. The diagonals of a parallelogram

are congruent.

6. An equiangular rhombus is a square.

7. Ifa polygon is a parallelogram,

it has four sides.

8. The diagonals of a quadrilateral

bisect each other.

9. A paraJlelogram is a rectangle.

10. The diagonals of a rhombus are

perpendicular to each other.

11. The measure of the line segment joining the midpoints of two sides of a triangle is equal to the measure of the third side.

12. If two angles have their

corresponding sides respectively parallel to each other, they are either

congruent or

sUpplementary.

13. If the diagonals of a parallelogram

are congruent,

the parallelogram is a rectangle.

14. If the diagonals of a parallelogram

are perpendicular, the parallelogram is a square.

15. A parallelogram

is defined as a quadrilateral

the opposite sides of which arc congruent.

16. If the diagonals of a quadrilateral

are perpendicular to each other, the quadrilateral is a parallelogram.

17. The

nonparaJlel sides of an isos,."L.

Summary Tests

Test 1

Copy the following chart and then place a check mark in the space provide, if the figure has the given property.

Roth pairs of

°PIY""

congruent.

IS. The line

^{..r nase.}

'egmen" joining Ihe

midpoint' of °PIMite ,id" of a qnad.

rilateral bisect each other.

19. If two sides of a quadrilateral are congruent, it is a parallelogram.

20. The

median of a trapezoid bisects each diagonal.

21. .fhe diago nab of a

pa'aHeJogra m divide it into four con g.u",t Itian gle,.

22. The line,

Ihmugh the vettice, of a pacaHeJogTam

pacaHeJ 10 Ihe diagonal, form another parallelogram.

23. If the

diagonal, of a "eetangle ace

Petpendicular, 'he pacaHeiogtam ISa square.

2'. 'rh,

'egment'joiniog the

con'<cntive ",idpuint, of the ,id" of a c<c'angle form a rhombus.

25 Th . .

h . .

d

' f .

d r . e segments JOlfling t e consecutIve ml POlflts 0 a trapezOl

lorm a parallelogram.

26. A

'Inadcila'"al i, a pataHeiogcam if

i" diagonal, ace peependicul.. 10 each other.

aRe congruent

angles Both pairs of opposite angles are

congruent.

Diagonals are of equal length.

Diagonals bisect each other.

Diagonals are perpendicular.

All sides are congruent.

All angles are congruent.

202

203

Test 3

204 FUNDAMENTALS OF COLLEGE GEOMETRY

27. A trapezoid is equilateral if it has two congruent sides.

28. The sum of the measures of the angles of a quadrilateral is 360.

29. The bisectors of the opposite angles of a rectangle are parallel.

30. The bisectors of the adjacent angles of a parallelogram are perpendicular.

PROBLEMS

I. AC is the diagonal of rhombus ABCD. If mLB

= 120, find mLBAC.

2. InD ABCD, mAB = 10 inches, mLB = 30, and AH ..1 BC. Find mAH.

3. InD ABCD, mLA = 2mLB. Find mLA.

4. In 0 ABCD, diagonal AC ..1 BC and AC ==

BC. Find mLD.

5. In LABC, AD

==

DB; mLC = 90; mLB = 30; mAC = 14 inches. Fin mBD.

6. In LABC, AD

==

DB; mLC = 90; mLA = 60; mCD = 12 inches. Fin' mAC.

7. In LABC, AD

==

DB; mLC = 90; mLA = 60; mAB = 26 inches.

mCD.

A

I

i .

c

2.

Given: AM is a median of LABC; LDCB

^~----