The straight lines AD and CD, projected to a point D within the triangle ADC from the extremities of the base AC, are together less than the sides AB and CB of the triangle, but contain a greater angle.

B

D

E

C

For produce AD to meet CB in E. The two sides AB and BE of the triangle ABE are greater than the third side AE (I. 16.); add EC to each, and AB, BE, EC, or AB and BC, are greater than AE and EC. But the sides CE and ED of the triangle DEC are (I. A 16.) greater than DC, and consequently CE, ED, together with DA, or CE and EA, are greater than CD and DA. Wherefore the sides AB and BC, being greater than AE and EC, which are themselves greater than AD and DC, must be much greater than AD and DC, or the lines AD and DC are much less than AB and BC the sides of the triangle.

Again, the angle ADC, being the exterior angle of the triangle DCE, is greater than DEC (I. 10.); and for the same reason, DEC is greater than ABE, the opposite interior angle of the triangle EAB. Consequently ADC is much greater than ABE or AВС.

PROP. XX. THEOR.

If two sides of one triangle be respectively equal to those of another, but contain a greater angle ; the base also of the former will be greater than that of the latter.

In the triangles ABC and DEF, let the sides AB and BC be equal to DE and EF, but the angle ABC greater than DEF; then is the base AC greater than DF.

For draw BG equal to EF and making an angle ABG equal to DEF (I. 4.), join AG and GC.

Because AB and BG are equal to DE and EF, and the contained angle ABG is equal to DEF; the triangles ABG

1

and DEF (I. 3.) are equal, and have equal bases AG and

which is greater than ACG, must be much greater than ACG; and therefore the opposite side AC is (I. 15.) greater than AG or DF.

PROP. XXI. THEOR.

If two sides of one triangle be respectively equal to those of another, but stand on a greater base; the angle contained by the former will be likewise greater than what is contained by the latter.

Let the triangles ABC and DEF have the sides AB and BC equal to DE and EF, but the base AC greater thran DF; the vertical angle ABC is greater than DEF.

From AC cut off AG equal to DF, construct (I. 1.) the triangle AHG having the sides AH and GH equal to AB and BC or DEA

and EF, join HB, and produce HG to meet BC in I.

H

B

I

CD

G

Because HI is greater than HG, it is greater than the equal side BC, and therefore much greater than BI. Consequently the opposite angle IBH of the triangle BIH is (I. 14.) greater than BHI. But AB being equal to AH the angle HBA is (1. 8.) equal to BHA, and therefore the two angles IBH and HBA are greater than IHB and BHA, that is, the whole angle CBA is greater than IHA or GHA. And since the sides of the triangle AGH are by construction equal to those of EDF, the corresponding angle AHG is equal to DEF (I. 2.); and hence the angle ABC, which is greater than AHG, is likewise greater than

DEF.

PROP. XXII. THEOR.

If straight lines be drawn from the same point to another straight line, the perpendicular is the shortest of them all; the lines equidistant from it on both sides are equal; and those more remote are greater than such as are nearer.

If the straight lines CG, CE, CD, and CF drawn from a given point C to the straight line AB, the perpendicular CD is the least, the equidistant lines CE and CF are equal, but the remoter line CG is greater than either of these two.

For the angle CDE, which is equal to CDF, is (I. 15.) greater than CFD, and consequently the opposite side CF is (I. 15.) greater than CD, or CD is less than CE.

C

AGE DF

B

But a straight line drawn of a determinate length from C to AB, may have two positions; for if the line CE be supposed to turn about the point C, the angle CEA will continually decrease (I. 13.), till, passing from obtuse to acute, it becomes equal to CEF, and then forms (I.9.) the isosceles triangle ECF.-Because ED is by hypothesis equal to FD, CD-common to the two triangles ECD and FCD, and the contained angles CDE and CDF equal; these triangles (I. 3.) are equal, and consequently their bases CE and CF are equal.

Again, because GCD is a right angled triangle, the angle CGD or CGE is acute (I. 12.), and for the same reason, the angle CED of the triangle CDE is acute, and consequently its adjacent angle CEG is obtuse. Wherefore CEG is much greater than CGE, and the opposite side CG greater (I. 15.) than CE. 27

Cor. Hence only a single perpendicular CD can be let fall from the same point Cupon a given straight line AB; and hence also a pair only of equal straight lines greater than CD can at once be extended from C to AB, making on the same side, the one an obtuse angle CEA, and the other an acute angle CFA. As the term distance signifies the shortest road, the distance between two points is the straight line which joins them, and the distance from a point to a straight line is the perpendicular let fall upon it.

PROP. XXIII. THEOR.

Two triangles are equal, which have two angles and a corresponding side in the one respectively equal to those in the other.

Let the triangles ABC and DEF have the angle BAC equal to EDF, the angle BCA to EFD, and a side of the one equal to a side of the other, whether it be interjacent or opposite to those equal angles; the triangles will be equal.

First, let the equal sides be AC and DF, which are interjacent to the equal angles in both triangles.-Apply the triangle ABC to DEF; the point A being laid on D, and the straight line AC on DF, the other extremities C and

E

F must coincide, since those lines are equal. And because the angle CAB is equal to FDE, and the side AC is

B

applied to DF, the other side AB must lie along DE; and for the same A reason, the angles BCA and EFD being equal, the side CB must lie along FE. Wherefore the point B, which is

CD

common to both the lines AB and CB, will be found likewise in both DE and FE; that is, it must fall upon the corresponding vertex E. The two triangles ABC and DEF, thus adapting, are hence entirely equal.

Next, let the equal sides be AB and DE, which are opposite to the equal angles ACB and DFE. The triangle ABC being laid on to DEF, the sides AB and AC of the angle BAC, will apply to DE and DF, the sides of the

equal angle EDF; and since AB is

equal to DE, the points B and E

B

must coincide, but BC must adapt

itself to EF, for the angles BCA and

EFD are equal, and (Cor. I. 13.) only one straight line

can be drawn from the same point E to make a given angle on the same side with a given straight line. Whence the triangles ABC, DEF are entirely coincident, and have

those sides equal which subtend equal angles.

PROP. XXIV. THEOR.

Two triangles are equal, which, being of the same affection, have two sides and a corresponding angle in the one equal to those in the other.

Let the triangles ABC and DEF have the side AB equal to DE, BC to EF, and the angles BAC, EDF, opposite to BC, EF, also equal; the triangles themselves are equal, if both of them be right angled, or acute, or obtuse.

For the triangle ABC being applied to DEF, the angle BAC will adapt itself to EDF, since they are equal; and the point B must coincide with E, because the side AB is equal to DE. But the other equal sides BC and EF now

stretching from the same point E to

wards DF, must likewise coincide;

B

for if the angle at Cor F be right, there can exist no more than one per- A

pendicular EF (Cor. I. 22.) and, in like manner, if this