of the line ADF; therefore such oblique lines as lie farthest from the perpendicular are longest. Hence,

Of all the lines which can be drawn to a given straight line from any point without that line;

1. The perpendicular is the shortest.

2. Of the oblique lines, any two drawn on different sides of the perpendicular and equally distant from it, are equal to each

other.

3. Of the rest, that which is the most distant from the perpendicular is the longest.

Cor. 1. The perpendicular measures the true distance of a point from a line, because it is the shortest distance.

Cor. 2. From the same point three equal straight lines cannot be drawn to the same straight line; for if there could, we should have two equal oblique lines on the same side of the perpendicular, which is impossible.

If from C, the middle point of the straight line AB, the line EF be drawn perpendicular to this straight line; then,

1. Each point of the perpendicular will be equally distant from the two extremities of AB.

2. Every point situated without the perpendicular will be unequally distant from those extremities.

First. Since we suppose AC=CB, the two oblique lines AD, DB, are equally distant from the perpendicular, and are therefore equal. (Prop. 15.) So, likewise, AE=EB, and AF=FB, and so on. Therefore every point in the perpendicular is equally distant from the extremities A and B.

A

F

E

B

Second. Let I be a point out of the perpendicular. If IA and IB be joined, one of those lines will cut the perpendicular in D; from which, drawing DB, we shall have DBDA. But the straight line IB is less than ID+DB, and ID+DB=ID+DA=IA; therefore IB<IA. In like manner it may be proved that every point out of the perpendicular is unequally distant from the extremities A and B. Hence,

If a perpendicular be drawn from the middle point of a straight line, each point of the perpendicular will be equally distant from the extremities of that line; and every point without the perpendicular, will be unequally distant from those extremities.

Two right-angled triangles ABC, DEF, are every way equal, when the hypothenuse and a side of the one are respectively equal to the hypothenuse and a side of the other.

then will the triangle ABC=def.

The equality of the triangles would be manifest, if the third sides BC and EF were equal. (Prop. 10.) But suppose these sides are not equal, and that BC is the greater. Take BG=EF; and join AG. The triangle ABG=DEF; for the angles B and E are equal, (Ax. 11,) the side AB=DE by hypothesis, and BG EF by construction; hence these triangles are every way equal, (Prop. 5,) consequently AG= DF. Now by hypothesis DF-AC; and therefore AG= AC. (Ax. 1.) But, (Prop. 15,) the oblique line AC cannot be equal to AG, which lies nearer the perpendicular AB;

therefore it is impossible that BC should differ from EF; hence the triangles ABC and DEF are equal. In a similar manner it may be shown that two right-angled triangles are equal when the base and hypothenuse of the one are respectively equal to the base and hypothenuse of the other. Hence,

Any two right-angled triangles are every way equal, when the hypothenuse and a side of the one, are respectively equal to the hypothenuse and a side of the other.

Scholium. The equality of triangles has been proved in four different cases, and in each case from different data; but all of them may be referred to the fifth, or the sixth proposition.

If two straight lines AC, BD, are perpendicular to a third AB, they will be parallel; in other words, they will never meet, how far soever both of them be produced.

straight line, which is impossible. (Prop. 14.)

In like manner it may be shown that AC, BD, will not meet, if produced in the opposite direction. Therefore they are parallel. (Def. 14.) Hence,

Any two straight lines, which are perpendicular to a third line, are parallel.

PROPOSITION XIX. LEMMA.

If the straight line BD is perpendicular to AB, and if another straight line AE makes with AB the acute angle BAE, then the straight lines BD, AE, if produced sufficiently, will

Again, let another point L be taken in the line AE, at a distance AL greater than AF, and let LM be drawn from it perpendicular to AB. It can be proved as above, that the point M must fall in the direction GB; so that the distance AM will, of necessity, be greater than AG.

Now it may be observed, that if AL be taken double of AF, AM will be double of AG; also, if AL is taken triple of AF, AM will be triple of AG; and, generally, there is al ways the same ratio between AM and AG, as between AL and AF. Therefore it follows that the straight line AE, if produced sufficiently, will meet BD. Hence,

If a straight line, meeting two other straight lines, makes one of the interior angles a right-angle, and the other an acute angle, those lines will meet if sufficiently produced.

Cor. The distance on the line AE, where the two lines will meet, may be determined by the following proportion, AG: AB: AF : x.

If two straight lines, AC, BD, form with a third AB, two interior angles CAB, ABD, whose sum is equal to two rightangles, the lines AC, BD, are parallel.

E

A

C

D

From G, the middle point of AB, draw the straight line EGF perpendicular to AC. It will also be perpendicular to BD. For the sum of the angles GAE+GBD is equal to two right-angles by hypothesis: the sum GBF+GBD is likewise equal to two right-angles; (Prop. 1;) hence GAE+GBD=GBF+GBD; taking GBD from both, there remains the angle GAE=GBF. Again, the angles AGE, BGF are equal; (Prop. 4;) therefore the triangles AGE and BGF have each a side and two adjacent angles equal; hence, (Prop. 6,) they are themselves equal, and the angle BFG is equal to AEG: but AEG is a right-angle by construction; therefore the lines AC, BD, being perpendicular to the same straight line EF, are parallel. (Prop. 18.) Hence,

Any two straight lines which, meeting a third straight line, make the sum of the two interior angles on the same side equal to two right-angles, are parallel.

If two straight lines AB, CD, meeting a third line EF, make the alternate angles BGH, GHC equal to each other; or make the exterior angle EGB, equal to the opposite interior angle GHD, these two lines are parallel.