EF and CD, is equal to the difference in direction between EF and AB; and therefore CD is in the same direction with AB; but every part of the same straight line is in the same direction (Def. 14. 1.), and therefore every part of AB is in the same direction with every part of CD; and because AB and CD are in the same direction, and GH falls on them, the difference in direction between GH and AB is equal to the difference in direction between GH and CD; and therefore the angles which GH makes with CD, are equal to the corresponding angles which it makes with AB. Therefore, if a straight line, falling on two other straight lines, makes the corresponding angles equal, any other straight line falling on these two straight lines, will also make the corresponding angles equal. Which was to be proved.

COR. I. If a straight line falling on two other straight lines in the same plane, makes the corresponding angles equal; those two straight lines are parallel.

COR. II. If a straight line falling on two other straight lines in the same plane, makes the two interior angles on the same side together equal to two right angles; those two straight lines are parallel.

PROP. III. THEOR.

If a straight line fall on two parallel lines, it makes the corresponding angles equal.

E

A

G

B

Let the straight line EF fall on the two parallel lines AB, CD; it makes the corresponding angles equal. Because AB is parallel to CD, it is in the same direction (Def. 1. 2.); and therefore the difference in direction between AB and EF is equal to the difference in direction between CD and EF, and therefore the angles which EF makes with AB, are equal to the corresponding angles which it makes with CD.

C

H

D

F

COR. I. Because the angle EGB is equal to the angle GHD, the angles EGB and BGH are together equal to the angles GHD and BGH; but EGB and BGH are together equal to two right angles; therefore GHD and BGH are together equal to two right angles. Therefore,

if a straight line fall on two parallel lines, it makes the two interior angles on the same side together equal to two right angles.

COR. II. Straight lines that are parallel to the same straight line, are parallel to one another.

PROP. IV. PROB.

To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw a straight line through the point A parallel to the straight line BC. In BC take any point D, and join AD; and at the point A in the straight line AD, make (14. 1.) the angle DAE equal to the angle ADC; and produce the straight line AE to F.

A

Because the straight line AD, falling on the two straight lines BC, EF, makes an angle DAE, with one of the lines EF, equal to an angle ADC, which it makes with the other line BC, the remaining angles which AD makes with EF, are equal to the remaining angles it makes with BC (1. 2.); and any other straight line falling on EF

E

B

D

F

C

and BC, will also make the corresponding angles equal (2. 2.); and therefore EF is in the same direction as BC, or is parallel to it. Therefore, the straight line EAF is drawn through the given point A, parallel to the given point BC. Which was to be done.

PROP. V. THEOR.

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

Let ABC be a triangle, and let any of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB, are together equal to two right angles.

Through the point C draw CE parallel (4. 2.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the corresponding angles BAC, ACE are equal (3. 2.). Again, because AB

is parallel to CE, and BD falls upon them, the corresponding angles ABC, ECD are equal; but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC: to these equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal (3 Cor. 1. 1.) to two right angles: therefore, also, the angles CBA, BAC, ACB, are equal to two right angles. Wherefore, if a side of a triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Which was to be proved.

COR. I. 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.

D

For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F which is the common vertex of the triangles; that is (Cor. 2. 3. 1.), together with four right angles. Therefore all the angles of the figure, together with

A

F

B

four right angles, are equal to twice as many right angles as the figure has sides.

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

Because every interior

angle ABC with its adjacent exterior ABD is equal (Cor. 3. 1. 1.) to two right angles; therefore all the interior together with all the exterior angles of the figure, are equal to twice

A

as many right angles as

there are sides of the fi- D

B

C

gure, that is, by the fore

going Corollary, they are

equal to all the interior an

gles of the figure, together with four right angles: therefore all the exterior angles are equal to four right angles.

PROP. VI. THEOR.

If two triangles have two angles of the one equal to two angles of the other, each to each; and one side equal to one side, the triangles shall be equivalent.

Let ABC, DEF be two triangles which have the angles ABC, BCA, equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD; also let BC be equal to EF, the triangles shall be equivalent.

Because the three angles in each of the triangles ABC, DEF, are equal to two right angles (5. 2.), they are equal to each other; take away the angles ABC, BCA, equal to the angles DEF, EFD, and the remaining angle BAC will be equal to the remaining angle EDF; and if the side AB be not equal to the side DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join CG; therefore, because BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF,

each to each; and the angle GBC is equal to the angle DEF; therefore the base CG is equal (4. 1.) to the base DF, and the triangle GBC is equivalent (4. 1.) to the triangle DEF; therefore the angle GCB is equal to the angle DFE: but DFE is, by the hypothesis, equal to the angle ACB; wherefore, also, the angle GCB is equal to the angle ACB, the less to the greater, which is impossible; therefore AB is not unequal to DE, that is, it is equal to it, and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; the base, therefore, AC, is equal (4. 1.) to the base DF, and the triangle ABC is equivalent to the triangle DEF. Therefore, if two triangles have two angles of one equal to two angles of the other, each to each; and one side equal to one side, the triangles shall be equivalent. Which was to be proved.

PROP. VII. THEOR.

If two of the sides of a quadrangle are equal and parallel, the remaining sides are also equal and parallel.

Let ABDC be a quadrangle, and let the sides AB and CD be equal and parallel; AC and BD are also equal and parallel. Join CB.

Because AB is parallel to CD, the corresponding angles ABC and BCD are equal (3. 2.), and the side AB is equal to CD by the hy

pothesis, and BC is common; therefore (4. 1.) the triangles ABC, BCD are equivalent, and the side AC is equal to the side BD, and the angle ACB to the angle CBD; and therefore (Cor. 1. 2. 2.) AC is parallel to BD.

PROP. VIII. THEOR.

If two of the sides of a quadrangle are equal to the opposite two, each to each, they are also parallel.

Let ABDC be a quadrangle, and let the side AB be equal to the side CD, and let AC be also equal to BD: AB shall be parallel to CD, and AC to BD. Join BC.

Then, because the two triangles ABC and BCD have the side AB

equal to the side CD, and AC to BD by the hypothesis, and the third side BC common (7. 1.), they are equivalent; and therefore the angle ABC is equal to the angle BCD; and therefore AB (Cor. 1. 2. 2.) is parallel to CD; and because the angle ACB is equal to the angle CBD, AC is parallel to BD. Q. E. D.

PROP. IX. THEOR.

The opposite sides and angles of a parallelogram are equal.

In the quadrangle ABDC, lct AB be parallel to CD, and AC to BD. AB is equal to CD, and AC to BD. Join BC.

Because AB is parallel to CD, and BC meets them, the corresponding angles ABC, BCD are equal (3. 2.) to one another; and because AC is parallel to BD, and BC meets them, the corresponding angles ACB, CBD are equal (3. 2.) to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two