BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB.

SCHOLIUM.

This may be demonstrated without producing any of the sides: thus, the line BC, for example, is the shortest distance from B to C; therefore BC is less than BA+AC, or BA+AC>BC.

PROP. XIV. THEOR.

If from the ends of one side of a triangle, there be drawn two straight lines to a point within the triangle, these two lines shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC.

A

E

D

Produce BD to E; and because two sides of a triangle (13. 1.) are greater than the third side, the two sides BA, AE of the triangle ABE are greater than BE. To each of these add EC; therefore the sides BA, AC are greater than BE, EC: Again, because the two sides CE, ED, of the triangle CED are greater than CD, if DB be added to each, the sides CE, EB, will be greater than CD, DB; but it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle (9. 1.) is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle ČEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC.

B

C

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides of the other; the base of that which has the greater angle shall be greater than the base of the other.

Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make the angle

EDG equal to the angle BAC: and make DG equal to AC or DF, and join EG, GF..

Because AB is equal

to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal to the angle EDG, therefore the base BC is equal (1. 1.) to the base EG; and because DG is equal to DF, the angle DFG is equal (3. 1.) to the angle DGF; but the angle DGF is

AA

greater than the angle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, and because the greater (12. 1.) side is opposite to the greater angle, the side EG is greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF.

PROP. XVI. THEOR.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides of the other.

Let ABC, DEF be two triangles which have the two sides, AB, AC, equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF: but let the base CB be greater than the base EF, the angle BAC is likewise greater than the angle EDF.

For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (1. 1.) to EF; but it is not; therefore the angle BAC is not equal to the angle EDF; neither is it less; because then the base BC would be less (15. 1.) than the base EF; but it is not; therefore the angle

BAC is not less than the angle

:

D

AA

B

E

F

EDF and it was shewn that it is not equal to it: therefore the angle BAC is greater than the angle EDF.

A perpendicular is the shortest line that can be drawn from a point, situated without a straight line, to that line: any two oblique lines drawn from the same point on different sides of the perpendicular, cutting off equal distances on the other line, will be equal; and any two other oblique lines, cutting off unequal distances, the one which lies farther from the perpendicular will be the longer.

If AB, AC, AD, &c. be lines drawn from the given point A, to the indefinite straight line DE, of which AB is perpendicular; then shall the perpendicular AB be less than AC, and AC less than AD, and so on.

For, the angle ABC being a right one, the angle ACB is acute, (Th. 10.) or less than the angle ABC. But the less angle of a triangle is subtended by the less side (Th. 12.); therefore, the side AB is less than the side AC.

Again, if BC=BE; then the two oblique lines AC, AE, are equal. For the side AB is common to the two triangles ABC, ABE, and the contained angles ABC and ABE equal; the two triangles must be equal (Th. 1.); hence AE, AC are equal.

D

C

A

B E

Finally, the angle ACB being acute, as before, the adjacent angle ACD will be obtuse; since (Th. 6.) these two angles are together equal to two right angles; and the angle ADC is acute, because the angle ARD is right; consequently, the angle ACD is greater than the angle ADC; and, since the greater side is opposite to the greater angle (Th. 12.); therefore, the side AD is greater than the side AC.

COR. 1. The perpendicular measures the true distance of a point from a line, because it is shorter than any other distance.

COR. 2. Hence, also, every point in a perpendicular at the middle point of a given straight line, is equally distant from the extremities of that line. COR. 3. 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.

PROP. XVIII. THEOR.

When the hypotenuse and one side of a right angled triangle, are respectively equal to the hypotenuse and one side of another; the two right angled triangles are equal.

Suppose the hypotenuse ACDF, and the side AB-DE; the rightangled triangle ABC will be equal to the right-angled triangle DEF.

Their equality would be manifest, if the third sides BC and EF were equal. If possible, suppose that those sides are not equal, and that BC is the greater. Take BH=EF; and join AH. The triangle ABH DEF;

=

for the right angles B and E are equal, the side AB=DE, and BH=EF; hence, these triangles are equal (Th. 1.), and consequently AH-DF. Now (by hyp.), we have DF AC; and therefore, AH= AC. But by the last proposition, the oblique line AC cannot be equal to the oblique line AH, which lies nearer to the perpendicular AB; therefore

it is impossible that BC can differ from EF: hence, then the triangles ABC and DEF are equal.

PROP. XIX. THEOR.

If a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines are parallel.

Let the straight line EF, which falls upon the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.

A

E

B

G

D

For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C ; let them be produced and meet towards B, D in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater (9. 1.) than the interior and opposite angle EFG; but it is also equal to it, which is impossible: therefore, AB and CD being produced, do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel (15 Def.) to one another. AB therefore is parallel to CD.

C

F

PROP. XX. THEOR.

If a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines are parallel to one another.

Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to GHD, the interior and opposite angle upon the same side; or let it make the interior angles on the same side BGH, GHD together equal to two right angles; AB is parallel to CD.

Because the angle EGB is equal to the angle GHD, and also (8. 1.) to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel (19. 1.) to CD. Again, because the angles BGH, GHD are equal (by Hyp.) to two right angles, and AGH, BGH, are also equal (6.1.) to two right angles, the angles AGH, BGH are equal to the angles BGH,

GHD: Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD.

COR. Hence, when two straight lines are perpendicular to a third line, they will be parallel to each other.

PROP. XXI. THEOR.

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

the same

Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles.

For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL will be parallel to CD (19. 1.); but AB is also parallel to CD ; therefore two straight lines are drawn through the same point G, parallel to CD, and yet not coinciding with one another, which is impossible (11. Ax.)

The angles AGH, GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGH (S. 1.); and AGH is proved to be equal to GHD; therefore EGB is likewise equal to GHD add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH, are equal (6. 1.) to two right angles; therefore also BGH, GHD are equal to two right angles.

COR. 1. If BGH is a right angle, GHD will be a right angle also ;: