and AD be taken from each sum, the remainder B C must be greater than the remainder D B; or D B will be less than B C. Q. E. D.

THEOREM XVII.

If a line EF intersect two parallel lines A B and CD, the alternate angles B E F and CFE will be equal to each other.

A E

F

B

For if they are not equal, one of them must be greater than the other. Suppose BEF to be the greater, and let DE F, a part of B E F, be equal to CFE. Then because E B is parallel to C F, ED which meets E B, will also if produced meet C F in some point, as D. Then because E FC is the outward angle of the triangle E FD, it is greater than the inward and opposite angle FED, (Theo. 9.); but these angles are also by supposition equal, which is impossible. The angles B E F and C F E are therefore not unequal, that is they are equal to each other. Q. E. D.

Cor. A line which is perpendicular to one of two parallel lines, is also perpendicular to the other.

THEOREM XVIII.

G

E B

If a line E F meet two other lines A B and C D, and the alternate angles B E F and E F C be equal, the lines A B and C D are parallel. For if C D is not parallel to A B, from F, the point in which E F meets C D, let F G be drawn parallel A to A B. Then as the angles BE F and E FC are. equal by hypothesis, and the alternate angles BEFC and E F G are also equal, (Theo. 17) the angle EFG must be equal to the angle EFC; the less to the greater, which is impossible. Therefore no line passing through F, except C D, is parallel to A B; and consequently CD is parallel to A B. Q. E. D.

Cor. Lines which are perpendicular to the same line are parallel to each other.

THEOREM XIX.

If two parallel lines A B, C D, be cut by another line E H, in the points F and G, any outward angle, as E F A, will be equal to the inward opposite angle F G C on the same side of E H; and any two inward angles, as A F G and F G C, on the same side of E H, will together be equal to two right angles.

For the angle AFE is equal to BF G, (Theo. 8.) which angle is equal to F G C, (Theo. 17.); hence the angles AF E and F G C are equal. AFE and A F G together make two right angles, (Theo. 6.); therefore A F G and F G C together are equal to two right angles. Q. E.D.

F

F

B

But

C

D

H

Cor. 1. If a line meeting two other lines make the outward angle equal to the inward and opposite one, on the same side of the line; or if the two inward angles on the same side of the line be, together, equal to two right angles, the alternate angles will be equal; and consequently these two lines will be parallel.

Cor. 2. If a line cut two other lines, and the outward angle be not equal to the inward and opposite one on the same side of the line; or if the two inward angles on the same side of the line, are not together equal to two right angles; these two lines are not parallel, and will therefore meet if produced.

THEOREM XX.

A

If any two lines, A B and C D, be each parallei to another line E F ; the lines A B and C D are parallel to each other. For let the line G L cut the lines A B, C D, and E F, in the points H, I, and K. Then the angles G H B and H I D, being each equal to IK F, (Theo. 19.) are equal to each other. But GH B is equal to A H I, (Theo. 8.) ; therefore A HI and HID are equal; and hence A B` and C D are parallel, (Theo. 18.)

Q. E. D.

THEOREM XXI.

G

B

I

D

E.

K

F

L

If two lines A C, A D meeting in a point A, be parallel to two other lines E B, BF, meeting in another point B; the included angles CAD and E B F are equal to each other.

B G

For join A B and produce the line to G, then, as the angle CA B is equal to the angle E B G, and the angle D A B to the angle F B G, C D (Theo. 19.); if the equal angles B A D and GBF, be taken from the equal angles B A C and G B E, the remaining angles C.AD and E BF will be equal. Q. E. D.

THEOREM XXII.

E F

If any side A B, of a triangle A B C, be produced, the outward angle CBD, is equal to both the inward and opposite angles B A C, and AC B, taken together.

For let B E be a line parallel to A C. Then the angle C will be equal to the angle C B E, (Theo. 17.); and the angle A to the angle E BD, (Theo. 19.), hence the whole angle C B D will be equal to the two angles

A

B

A and C together. Q. E.D.

Cor. The angle C is the difference between the outward angle CBD and the other inward and opposite angle A; and A is the difference between C B D and C.

THEOREM XXIII.

In any triangle ABC, if one of the sides, as A B, be produced, and the outward angle C B D, and one of the inward and opposite angles, as BAC, be bisected by the lines B E and A E, the angle E formed by the bisecting lines, is equal to half the angle A CD, the remaining inward and opposite angle of the triangle.

C

B D

E

For the angle C B D is equal to the angles B A C and A C B together, (Theo. 22.); therefore EBD, which is the half of C B D, is equal to the half of the angles B A C and B C A together, or to the angle BA E and the half of B CA. But E B D is A also equal to the angles BAE and E together, (Theo. 22.) Therefore the angles B A E and E together, are equal to the angle BAE and half the angle C together; and hence, by omitting the common angle BAE, we have the angle E equal to half of BCA. Q. E. D.

THEOREM XXIV,

The three interior angles of any triangle A B C, are together equal to two right angles.

For let A B be produced to D; then the angle CBD is equal to the angles A and C together (Theo. 22.) But the angles C B D and C B A are together equal to two right angles, (Theo. 6.); whence the three interior angles A B C, A C B, and B A C are A together equal to two right angles. Q. E. D.

B D

Cor. 1. If two angles in one triangle be equal to two angles in another triangle, the remaining angles of those triangles are equal. Cor. 2. If one angle in a triangle be equal to an angle in another triangle, the sums of the remaining angles in each triangle will be equal.

Cor. 3. If one angle in a triangle be a right angle, the other two angles, together, will be equal to a right angle; and each of them will therefore be an acute angle.

Cor. 4. Every triangle has, at least, two acute angles.

THEOREM XXV.

All the inward angles of any rectilineal figure ABCD E, are together equal to twice as many right angles as the figure has sides, wanting four right angles.

D

C

For from any point, F, within the figure, let lines be drawn to its angular points, dividing the figure into as many triangles as it has sides. Then, the E interior angles of each triangle being together equal to two right angles, (Theo. 24.) the interior angles of all the triangles will, together, be_equal to twice as many right

A

B

angles as the figure has sides. But the interior angles of all the triangles, are the inward angles of the figure, and the angles about the point F; and the angles about the point F are four right angles, (Cor. 2. Theo. 6.) Hence, the interior angles of the figure and four right angles, are together equal to twice as many right angles as the figure has sides; or the interior angles of the figure, themselves, are equal to twice as many right angles as the figure has sides, wanting four right angles. Q. E. D.

Cor. 1. All the interior angles of any quadrilateral figure are together equal to four right angles.

Cor. 2. If the sum of two angles of a quadrilateral figure be equal to two right angles, the sum of the remaining angles will also be equal to two right angles,

THEOREM XXVI.

If the sides A B, BC, CD, &c. of any rectilineal figure be produced, the outward, or exterior angles A, B, C, &c. will together be equal to four right angles.

D

a

A

B

For let the inward angles be denoted, as in the figure, by the letters a, b, c, &c. then each exterior angle with its corresponding interior one, as A and a, B and b, C and c, together make two right angles; therefore (Theo. 6.) all the exterior and all the interior angles will together make twice as many right angles as the figure has sides. But all the interior angles, and four right angles, are also together equal to twice as many right angles as the figure has sides, (Theo. 25.) Hence the interior and the exterior angles of the figure are, together, equal to the interior angles of it and four right angles. If, therefore, the interior angles be omitted from each sum, the exterior angles will remain equal to four right angles. Q. E. D.

THEOREM XXVII.

If from a point A, a perpendicular A B be drawn to any line D E, this perpendicular is the shortest line that can be drawn from the point A to the line DE; and of other lines as A C and AD, drawn from the same point A to the line DE, that which is nearest the perpendicular is less than one more remote.

A

For the angle ABC being a right angle, the angle ACB is less than a right angle, (Cor. 3. Theo. 24.); hence the side AB is less than A C, (Theo. 12.) Again, because A C B is less than a right angle, A C D is greater than a right angle, (Theo. 6.); and consequently the angle A D C is less than a right angle, (Theo. 24.); and therefore the side A C is less than the side A D, (Theo. 12.) Q. E. D.

D C B E

THEOREM XXVIII.

In any parallelogram A B C D, the opposite sides are equal to each other, and so also are the opposite angles; and the diagonal B D divides it into two equal triangles.

C

B

For since A B and D C are parallel, the angles A B D and C D B are equal, (Theo. 17.); and for a like reason the angles A D B and C B D are equal. Therefore the triangles BD C and D B A are identical, (Theo. 2.); having two angles in the one equal to two angles in the other, and the side B D between the equal angles, common to both; the triangle A B D is therefore equal to the triangle BDC, the side A B to the side D C, A D to B C, and the angle A to the angle C; and as the two parts of the angle A D C are equal to the two corresponding parts of the angle A B C, the whole angles AD C and A B C are also equal. Q.E. D.

Cor. If one angle of a parallelogram be a right angle, all its angles are right angles, and consequently all the angles of a rectangle are right angles.

THEOREM XXIX.

Any quadrilateral, A B C D, whose opposite sides A B, C D, and AD, B C are respectively equal, is a parallelogram.

A

B

For let the diagonal B D be drawn; then as the triangles BA D, D C B, have the sides A B and DC equal, and also A D equal to B C, and the side B D common to both the triangles, they are identical, (Theo. 5.) ; therefore the angles B D C and A B D are equal, and so are the angles A D B and C B D. Hence A B is parallel to C D, and A D to BC, (Theo. 18.); and the figure is therefore a parallelogram. Q. E. D.

Cor. A square is a parallelogram, and hence also all the angles of a square are right angles.

THEOREM XXX.

The lines A C and BD, which join the corresponding extremities A and C, B and D, of the equal and parallel lines A B and C D, are themselves equal and parallel.

For let the points C and B be joined; then because C

A B and C D are parallel, the angles A B C and

B C D are equal, (Theo. 17.); and therefore, as A B

is equal to C D, and B C common to both triangles, they are equal in all respects, (Theo. 1.); consequently A C is equal to B D, and the angle AC B to the alternate angle D B C; and therefore A C and B D are parallel, (Theo. 17.) Q. E. D.