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PROP. XXVII. THEOR.

If a right line fall upon two other right lines, and make the alternate angles equal to one another, the two lines are parallel.

Let the line EF fall upon the two lines A B and C D, and make the alternate angles AGF and A EHD equal; then A B and C D are parallel.

For, if AB and C D be not parallel, they would meet if

с

F

E

B

H

D

produced. Suppose them to meet in L; then GHL is a triangle, and the exterior angle AG H is greater than the interior angle EHD (Prop. 16.); but these angles are also equal (Hyp.), which is absurd; therefore A B and C D do not meet if produced, and are therefore parallel (Def. 30.). Which was to be demonstrated.

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If a right line fall upon two other right lines, and make an exterior angle equal to the interior and opposite angle at the same side of the line or make the two interior angles at the same side equal to two right angles; the two lines are parallel.

Let the line EF fall upon the two lines A B and C D, and make the exterior angle EG B equal to the interior opposite angle EHD at the same side of EF: or make the two interior angles BGF and EHD at the same side equal to two right angles; then A B and C D are parallel.

A

E

B

G

H

D

F

For the angle AGF is equal to EGB (Prop. 15.), and the angle EGB is equal to EHD (Hyp.), therefore the angles AGF and E HD are equal (Ax. 1.); but they are alternate angles, therefore A B and C D are parallel (Prop. 27.).

Again: The angles BGF and AGF are equal to two right angles (Prop. 13.); but BGF and EHD are also equal to two right angles (Hyp.), therefore the angles BGF and AGF are equal to the angles B GF and EHD; take away the angle BGF common to both, and the remaining angles AGF and EHD are equal (Ax. 3.); but they are alternate angles, therefore A B and C D are parallel. W. W. D.

PROP. XXIX. THEOR.

If a right line fall upon two parallel right lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior opposite angle at the same side; and also the two interior angles at the same side, equal to two right angles.

Let the line EF fall upon two parallel lines AB and CD; then the alternate angles AGF and EHD are equal: and the exterior angle EGB equal to the interior angle EHD at the same side of E F and also the two interior angles BGF and EHD at the same side equal to two right angles.

A

E

B

G

H

D

F

Part 1. The alternate angles A G F and E HD are equal. For, if the angles AGF and E HD be not equal, one of them must be greater than the other; suppose AGF to be greater, and add the angle BGF to each, then the angles A G F and BGF together are greater than the angles E HD and BGF; but AGF and B G F are equal to two right angles (Prop. 13.); therefore EHD and B G F are less than two right angles, and therefore the lines A B and C D, if produced, would meet at the side BD (Ax. 12.): but they are parallel, and therefore cannot meet (Def. 30.), which is absurd; therefore neither of the angles A G F and E H D is greater than the other, therefore they are equal.

Part 2. The exterior angle EG B is equal to the interior opposite angle E HD. For the angles AGF and E GB are equal, being vertical opposite angles (Prop. 15.), and the angles AGF and E HD are equal, being alternate angles (Part 1.); since then each of the angles EGB and EHD is equal to the same angle AGF; they are equal to one another (Ax. 1.); therefore the angles E G B and EHD are equal.

Part 3. The two interior angles BGF and EHD are equal to two right angles. For the angles B G F and AGF are equal to two right angles (Prop. 13.); but AGF is equal to EHD, being alternate angles (Part 1.); therefore the angles BGF and EHD are equal to two right angles. Which was to be

demonstrated.

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If two right lines be parallel to the same right line, they are parallel to one another.

Let the lines A B and C D be parallel to the same line LN; then A B and C D are parallel to one another.

A

For the angles AGF and EPN are equal, being alternate angles (Prop. L 29.); but the angle EPN is equal to the interior opposite angle EHD (Prop. 29.); therefore the angles AGF and EHD are equal (Ax. 1.), and they are

C

F

E

B

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alternate angles; therefore the lines A B and CD are parallel (Prop. 27.). Which was to be demonstrated.

SCHOL. This proposition may be demonstrated without the aid of the 27th, thus: - Each of the angles EGB and E HD is equal to the same angle EPN (Prop. 29.); therefore they are equal to one another (Ax. 1.); and therefore the lines AB and CD are parallel (Prop. 28.).

COR. Two lines parallel to the same line, cannot pass through the same point.

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To draw a right line parallel to a given right line, through a given point.

Let A B be the given line, and c the given point; it is required to draw through c, a line parallel to a B.

In the line AB take any point D; join c D, and with it at the point c make an angle DCE equal to, CDB

E

A

c

(Prop. 23.), but at the opposite side of CD; then the line EF is parallel to A B. For, as the line CD falls upon the two lines A B and E F, and makes the alternate angles ECD and BDC equal to one another (Con.); therefore the two lines A B and EF are parallel (Prop. 27.). Which was to be done.

PROP. XXXII. THEOR.

If any side of a triangle be produced, the exterior angle is equal to the two interior opposite angles: and the three interior angles of every triangle, are equal to two right angles.

In the triangle ABC, let the side A B be produced to D; then the exterior angle CBD is equal to the two interior opposite angles CAB, and ACB and the three interior A angles CAB, ACB, and CBA together, are equal to two right angles.

C

B

D

From the point B draw B E parallel to AC (Prop. 31.): then, because the line BC falls upon the parallels AC and BC, the alternate angles ACB and EBC are equal (Prop. 29.); also, because AD falls upon the same parallels, the angle EBD is equal to the interior angle CAB (Prop. 29.); therefore the whole exterior angle CBD is equal to the interior opposite angles C A B and A C B.

Again: Because the angles CBA and CBD are equal to two right angles (Prop. 13.), but CBD was shown to be equal to the two interior angles CAB and ACB; therefore the three interior angles, CBA, CAB, and ACB, are equal to two right angles. W. W. D.

COR. 1. If one angle of a triangle be equal to the other two angles, it is a right angle.

COR. 2. A perpendicular may be drawn from the end of a line, by describing an equilateral triangle upon it as a base, and producing one of the sides till the part produced be equal to the base, and joining the end of the produced line to the end of the base.

COR. 3. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

COR. 4. All the exterior angles of any rectilineal figure are together equal to four right angles.

COR. 5. Each angle of an equilateral triangle is equal to a third of two right angles, or two-thirds of one right angle.

COR. 6. A right angle may be trisected, by describing an equilateral triangle on one of its sides, and bisecting that angle of the triangle which is at the vertex of the right angle.

PROP. XXXIII. THEOR.

The right lines which join the adjacent extremities of two equal and parallel lines, are themselves equal and parallel.

Let AB and CD be two equal and parallel lines; the right lines AC and BD joining the adjacent extremities a to c, and в to D, are also equal and parallel.

A

B

D

Draw the diagonal в C. And because the line B c falls on the parallels A B and CD, the alternate angles BCD and A B C are equal (Prop. 29.); then in the triangles A B C and B CD, the side A B is equal to CD, and BC is common to both, and the angle A B C is equal to B CD; therefore the side a c is equal to BD (Prop. 4.), and the angle ACB is equal to DBC; but they are alternate angles, therefore the lines A c and BD are parallel (Prop. 27.). W. W. D.

PROP. XXXIV. THEOR.

A

The opposite sides and angles of a parallelogram are equal to one another, and the diagonal divides it into two equal parts. Let ABCD be a parallelogram, and B C the diagonal; the side AB is equal to CD, and AC to BD; the angles at A and D are equal, also the angles at B and c are equal; and the diagonal BC divides the parallelogram into two equal parts.

B

D

Because the line вc falls on the parallels A B and CD, the alternate angles ABC and BCD are equal (Prop. 29.); also because BC falls on the parallels AC and BD, the alternate angles ACB and DBC are equal (Prop. 29.); then in the triangles ABC and BCD, the angles A B C and ACB in the one, are equal to the angles BCD and DBC in the other, and the side BC is common to both; therefore the two triangles are equal in all respects (Prop. 26.), the side A B is equal to CD, and AC to B D, and the angles at A and D are equal; also the two angles ACB and BCD are respectively equal to the two angles DBC and ABC (Prop. 26.); therefore the whole angle ACD is equal to the whole angle ABD. And as the triangles ABC and BCD are equal, the parallelogram is divided into two equal parts by the diagonal B C. W. W. D.

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