lines AC, CF is less than the sum of the two lines AD, DF. Therefore, AC, the half of ACF, is less than AD, the half of ADF; hence the oblique line which is furthest from the perpendicular is the longest. Therefore, if from a point, &c.

Cor. 1. The perpendicular measures the shortest distance of a point from a line, because it is shorter than any oblique

line.

Cor. 2. It is impossible to draw three equal straight lines from the same point to a given straight line.

PROPOSITION XVIII. THEOREM.

If through the middle point of a straight line a perpendicular is drawn to this line:

1st. Each point in the perpendicular is equally distant from the two extremities of the line.

2d. Any point out of the perpendicular is unequally dis tant from those extremities.

Let the straight line EF be drawn perpenlicular to AB through its middle point, C.

F

D

B

E

First. Every point of EF is equally distant from the extremities of the line AB; for, since AC is equal to CB, the two oblique lines AD, DB are equally distant from the A perpendicular, and are, therefore, equal (Prop. XVII.). So, also, the two oblique lines AE, EB are equal, and the oblique lines AF, FB are equal; therefore, every point of the perpendicular is equally distant from the extremities A and B. Secondly. Let I be any point out of the perpendicular. Draw the straight lines IA, IB; one of these lines must cut the perpendicular in some point, as D. Join DB; then, by the first case, AD is equal to DB. To each of these equals add ID, then will IA be equal to the sum of ID and DB. Now, in the triangle IDB, IB is less than the sum of ID and DB (Prop. VIII.); it is, therefore, less than IA; hence, every point out of the perpendicular is unequally distant from the extremities A and B. Therefore, if through the middle point, &c.

Cor. If a straight line have two points, each of which is equally distant from the extremities of a second line, it will be perpendicular to the second line at its middle point.

B

If two right-angled triangles have the hypothenuse and a side of the one, equal to the hypothenuse and a side of the other each to each, the triangles are equal.

A

D

Let ABC, DEF be two right-angled triangles, having the hypothenuse AC and the side AB of the one, equal to the hypothenuse DF and side DE of the other; then will the side BC be equal to EF, and the triangle ABC to the triangle DEF.

B

G CE

F

For if BC is not equal to EF, one of them must be greater than the other. Let BC be the greater, and from it cut off BG equal to EF the less, and join AG. Then, in the triangles ABG, DEF, because AB is equal to DE, BG is equal to EF, and the angle B equal to the angle E, both of them being right angles, the two triangles are equal (Prop. VI.), and AĞ is equal to DF. But, by hypothesis, AC is equal to DF, and therefore AG is equal to AC. Now the oblique line AC, be ing further from the perpendicular than AG, is the longer (Prop. XVII.), and it has been proved to be equal, which is impossible. Hence BC is not unequal to EF, that is, it is equal to it; and the triangle ABC is equal to the triangle DEF (Prop. XV.). Therefore, if two right-angled triangles, &c.

PROPOSITION XX. THEOREM.

Two straight lines perpendicular to a third line, are paruilel.

Let the two straight lines AC, BD be both perpendicular to AB; then is AC parallel to BD.

For if these lines are not parallel, being produced, they

must meet on one side or the other of AB. Let them be produced, and meet in O; then there will be two perpendicu lars, OA, OB, let fall from the same point, on the same straight line, which is impossible (Prop. XVI.) Therefore, two straight lines, &c.

PROPOSITION XXI. THEOREM.

If a straight line, meeting two other straight lines, makes the interior angles on the same side, together equal to two right angles, the two lines are parallel.

Let the straight line AB, which A meets the two straight lines AC, BD, make the interior angles on the same side, BAC, ABD, together equal to two right angles; then is AC parallel to BD.

E

G

T B

C

From G, the middle point of the line AB, draw EGF perpendicular to AC; it will also be perpendicular to BD. For the sum of the angles ABD and ABF is equal to two right angles (Prop. II.); and by hypothesis the sum of the angles ABD and BAC is equal to two right angles. Therefore, the sum of ABD and ABF is equal to the sum of ABD and BAC. Take away the common angle ABD, and the remainder, ABF, is equal to BAC; that is GBF is equal to GAE.

Again, the angle BGF is equal to the angle AGE (Prop. V.); and, by construction, BG is equal to GA; hence the triangles BGF, AGE have two angles and the included side of the one, equal to two angles and the included side of the other; they are, therefore, equal (Prop. VII.); and the angle BFG is equal to the angle AEG. But AEG is, by construction, a right angle, whence BFG is also a right angle; that is, the two straight lines EC, FD are perpendicular to the same straight line, and are consequently parallel (Prop. XX.). Therefore, if a straight line, &c.

Scholium. When a straight line intersects two parallel lines, the interior angles on the same side, are those which lie within the parallels, Aand on the same side of the secant ine, as AGH, GHC; also, BGH, GHD.

Alternate angles lie within the parallels, on different sides of the

H

F

E

-B

secant line, and are not adjacent to each other, as AGH GHD; also, BGH, GHC.

Either angle without the parallels being called an exterior angle, the interior and opposite angle on the same side, lies within the parallels, on the same side of the secant line, but

not adjacent, thus, GHD is an interior angle opposite to the exterior angle EGB; so, also, with the angles CĤG, AGE.

If a straight line, intersecting two other straight lines, makes the alternate angles equal to each other, or makes an exterior ingle equal to the interior and opposite upon the same side of the secant line, these two lines are parallel.

C

G

E

B

Let the straight line EF, which intersects the two straight lines AB, CD, make the alternate angles AGH, GHD equal to each other; then AB Ais parallel to CD. For, to each of the equal angles AGH, GHD, add the angle HGB; then the sum of AGH and HGB will be equal to the sum of GHD and HGB. But AGH and HGB are equal to two right angles (Prop. II.); therefore, GHD and HGB are equal to two right angles; and hence AB is parallel to CD (Prop. XXI.).

D

H

F

Again, if the exterior angle EGB is equal to the interior and opposite angle GHD, then is AB parallel to CD. For, the angle AGH is equal to the angle EGB (Prop. V.); and, by supposition, EGB is equal to GHD; therefore the angle AGH is equal to the angle GHD, and they are alternate angles; hence, by the first part of the proposition, AB is parallel to CD. Therefore, if a straight line, &c.

PROPOSITION XXIII. THEOREM.

(Converse of Propositions XXI. and XXII.)

If a straight line intersect two parallel lines, it makes the alternate angles equal to each other; also, any exterior angle equal to the interior and opposite on the same side; and the two interior angles on the same side together equal to two right angles.

Let the straight line EF intersect the two parallel lines AB, CD; the alternate angles AGH, GHD are equal to each other; the exterior angle EGB is equal to the interior and opposite angle on the same side, GHD; and the two interior angles on the same side, BGH, GHD, are together equal to two right angle

K

E

G

L

B

H

C

D

F

For if AGH is not equal to GHD, through G draw the line KL, making the angle KGH equal to GHD; then KL must be parallel to CD (Prop. XXII.). But, by supposition, AB is parallel to CD; therefore, through the same point, G, two straight lines have been drawn parallel to CD, which is impossible (Axiom 12). Therefore, the angles AGH, GHD are not unequal, that is, they are equal to each other. Now the angle AGH is equal to EGB (Prop. V.), and AGH has been proved equal to GHD; therefore, EGB is also equa to GHD. Add to each of these equals the angle BGH; then will the sum of EGB, BGH be equal to the sum of BGH, GHD. But EGB, BGH are equal to two right angles (Prop. II.); therefore, also, BGH. GHD are equal to two right an gles. Therefore, if a straight line, &c

Cor. 1. If a straight line is perpendicular to one of two parallel lines, it is also perpendicular to the other.

Cor. 2. If two lines, KL and CD, make with EF the two angles KGH, GHC together less than two right angles, then will KL and CD meet, if sufficiently produced.

For if they do not meet, they are parallel (Def. 12). But they are not parallel; for then the angles KGH, GHC would be equal to two right angles.

Straight lines which are parallel to the same line, are paral

lel to each other.

Let the straight lines AB, CD be each of them parallel to the line EF; E. then will AB be parallel to CD.

For, draw any straight line, as C PQR, perpendicular to EF. Then, since AB is parallel to EF, PR, which Ais perpendicular to EF, will also be

perpendicular to AB (Prop. XXIII., Cor. 1); and since CD is parallel to EF, PR will also be perpendicular to CD. Hence, AB and CD are both perpendicular to the same straight line, and are consequently parallel (Prop. XX.). Therefore, straight lines which are parallel, &c.

Two parallel straight lines are every where equally distant from each other.

Let AB, CD be two parallel straight lines. From any