Because ADB is the exterior angle of the triangle BDC, it is greater (I._16) than the interior and opposite angle DCB. But the angle ADB is equal (I. 5) to the angle ABD, because the side AB is equal (Const.) to the side AD. Therefore the angle ABD is likewise greater than the angle A CB. Much more, therefore, is the angle ABC greater than the angle ACB. B Therefore the greater side, &c. QE. D.

D

PROP. XIX. (THEOREM.)-If two angles of a triangle (BAC) are unequal, the side opposite to or subtending the greater angle (B) is greater than the side opposite the less angle (C).

For, if the side AC be not greater than the side AB, AC must either be equal to A B, or less than it. The side AC is not equal to the side AB; because then the angle ABC would be equal (I. 5) to the angle ACB. it is (Hyp.) not. Therefore AC is not equal to

But

AB. The side AC is not less than the side AB. Because then the angle ABC would be less (I. 18) than the angle ACB. But it is (Hyp.)

not.

Therefore the side AC is not less than AB. And it has been shown that AC is not equal to AB. fore the side AC is greater than the side A. B. angle, &c. Q. E.D.

C

There

Wherefore the greater

PROP. XX. (THEOREM.)-Any two sides of a triangle (ABC) are together greater than the third side.

The sides BA, AC, are greater than the side BC. Produce BA to the point D, make (I. 3) AD equal to A C, and join DC. Then because DA is equal to AC, the angle ADC is equal (I. 5) to the angle ACD. But the angle BCD is greater (Ax. 9) than the angle ACD: therefore the angle BCD is greater than the angle ADC. But because the angle BCD of the triangle DCB is greater than its angle BDC, and the greater (I. 19) angle is subtended by the greater side. Therefore the side DB is greater than the side BC. Again AD is equal (Const.) to AC. To each of these

B

equals add BA. Therefore the whole BD is equal (Ax. 2) to the two BA and AC. But BD was proved to be greater than BC. Therefore the sides BA, AC, are greater than BC. In the same manner it may be demonstrated that the two sides AB, BC, are greater than ČA, and the two sides BC, CA, greater than AB. Therefore any two sides, &c. Q.E.D.

PROP. XXI. (THEOREM.)—If from the ends (B and C) of one side (B C) of a triangle (ABC), there be drawn two straight lines to a point (D) within the triangle, these (BD and CD) together shall be less than the other two sides of the triangle, but shall contain a greater angle.

Produce BD to E. Then, the two sides BA, AE, of the triangle

ABE are greater than the third side BE (I. 20).
unequals add EC. Therefore the two sides BA,
(Ax. 4) than BE, EC. Again, the two sides CE,
ED, of the triangle CED are greater (I. 20) than
the third side CD. To each of these unequals,
add DB. Therefore the two sides CE, EB, are
greater (Ax. 4) than CD, DB. But it has been
shown that BA, AC are greater than BE, EC.
Much more then are BA, AC, greater than Б
BD, DC.

To each of these
AC, are greater

Again, the exterior angle BDC of the triangle CDE is greater than its interior and opposite angle CED (I. 16). And the exterior angle CEB of the triangle ABE is greater than its interior and opposite angle BAC (I. 16). But the angle BDC is greater than the angle CEB. Much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q.E.D.

PROP. XXII. (PROBLEM).—To make a triangle of which the sides shall be equal to three given straight lines (A, B, and C), of which any two are greater than the third (I. 20).

D

Ꮐ

HE

LA

B

Take a straight line DE terminated at the point D, but unlimited towards E. Make (I. 3) DF equal to A, FG equal to B, and GH equal to C. From the centre F, at the distance FD, describe (Post. 3) the circle DKL. From the centre G, at the distance GH, describe (Post. 3) another circle HLK. And join KF, KG. The triangle KFG has its sides equal to the three straight lines A, B, C.

Because the point F is the centre of the circle DKL, FD is equal (Def. 15) to FK. But FD is equal (Const.) to the straight line A. Therefore FK is equal (Ax. 1) to A. Again, because G is the centre of the circle LKH, GH is equal (Def. 15) to GK. But GH is equal to C. Therefore also G K is equal to C. And FG is equal (Const.) to B. Therefore the three straight lines KF, FG, GK, are equal to the three straight lines A, B, C. Wherefore the triangle KFG has been made, having its three sides KF, FG, GK, equal to the three given straight lines A, B, C. Q. E. F.

с

A

PROP. XXIII. (PROBLEM).—At a given point (A) in a given straight line (AB), to make a rectilineal angle equal to a given rectilineal angle (D C È). In CD, CE, take any points D, E, and join DE. Upon the straight line AB make (I. 22) the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, that is, AF equal to CD, AG to CE, and FG to DE. The angle FAG is equal to the angle DCE.

Because the two sides FA, AG, are

E

G

D

F

B

equal to the two sides DC, CE, each to each, and the base FG to the base DE; therefore the angle FAG is equal (I. 8) to the angle DCE. Wherefore at the given point A, in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q. E. F.

PROP. XXIV. (THEOREM.)-If two triangles (ABC, DEF) have two sides (A B, A C) of the one equal to two sides (DE, EF) of the other, each to each, but the angle (BA C) contained by the two sides of one of them greater than the angle (EDF) contained by the two sides equal to them, of the other; the base (BC) of that which has the greater angle is greater than the base (E F) of the other.

Of the two sides DE, DF, let DE be the side which is not greater than the other. At the point D, in the straight line DE, make (I. 23) the angle EDG equal to the angle BA C. Make DG equal (I. 3) to AC or DF. And join EG, GF.

A

B

D

F

Because DE is equal (Hyp.) to AB, and DG (Const.) to AC, the two sides, ED, DG, are equal to the two BA, AC, each to each. And the angle EDG is equal (Const.) to the angle BAC. Therefore the base EG is equal (I. 4) to the base BC. Again, because DG is equal to DF, the angle DFG is equal (I. 5) to the angle DGF. But the angle DGF is greater (Ax. 9) than the angle EGF. Therefore the angle DFG is also greater than EGF. Much more then is the angle EFG greater than the angle EGF. Now, because the angle EFG of the triangle EFG is greater than its angle EGF, and the greater (I. 19) angle is subtended by the greater side; therefore the side EG is greater than the side EF. But EG was proved to be equal to BC. Therefore BC is greater than EF. Therefore if two triangles, &c. Q. E. D.

PROP. XXV. (THEOREM.)-If two triangles (ABC, DEF) have two sides (AB, AC) of the one respectively equal to two sides (DE, DF) of the other, but the base (BC) of the one greater than the base (EF) of the other; the angle (BAC) contained by the two sides of that which has the greater base, is greater than the angle (EDF) contained by the two sides equal to them of the other.

A

D

For, if the angle BAC be not greater than the angle EDF, it must either be equal to, or less than, the angle EDF. The angle BAC is not equal to the angle EDF, because then the base BC would be equal (I. 4) to the base EF: but it is (Hyp.) not equal. Therefore the angle BAC is not equal to the angle EDF. Again, the angle BAC is not less than the angle EDF, because then the base BC would be less (I. 24) than the base EF: but it is (Hyp.) not less. Therefore the angle BAC is not less than the angle EDF. And it was shown than the angle BAC is not equal to the angle EDF. Therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E.D.

B

CE

PROP. XXVI. (THEOREM.)-If two triangles (ABC, DEF) have two angles (ABC, BCA) of the one respectively equal to two angles (DEF, EFD) of the other, and one side equal to one side, viz., either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then their other sides are equal, each to each, and also the third angle of the one to the third angle of the other.

First, let those sides be equal which are adjacent to the angles that are equal in the two triangles, viz., BC to EF. Then their other sides are equal, each to each, viz., AB to DE, and

AC to DF; and the third angle BAC is equal to the third angle EDF.

For, if AB be not equal to DE one of them must be greater than the other. Let AB be the greater of the two. Make BG equal (I. 3) to DE, and join GC.

A

B

D

C E

F

Because in the two triangles GBC, DEF, BG is equal (Const.) to DE, and BC (Hyp.) to EF, the two sides GB, BC, are equal to the two sides DE, EF, each to each. But the angle GBC is equal (Hyp.) to the angle DEF. Therefore the base GC is equal (I. 4) to the base DF, and the triangle GBC to the triangle DEF. And the remaining angles of the one are equal to the remaining angles of the other, each to each, viz., those to which the equal sides are opposite. Therefore the angle GCB is equal to the angle DFE. But the angle DFE is (Hyp.) equal to the angle BCA. Wherefore also the angle BCG is equal (4x. 1) to the angle BCA, the less to the greater, which is impossible. Therefore the side AB is not unequal to the side DE; that is, AB is equal to DE. And BC is equal (Hyp.) to EF. Therefore the two sides AB, BC, are equal to the two sides DE, EF, each to each. And the angle ABC is equal (Hyp.) to the angle DEF. Therefore the base AC is equal (I. 4) to the base DF, and the third angle BAC to the third angle EDF.

Next, let those sides which are opposite to equal angles in each triangle be equal to one another, viz.. AB to DE. Then their other sides are equal, viz., AC to DF, and BC

to EF. And the third angle BAC is A equal to the third angle EDF.

B

D

EC E

F

For, if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater of the two. Make BH equal (I. 3) to EF, and join A H. Because in the two triangles ABH, DEF, BH is equal (Const.) to EF, and AB to (Hyp.) DE; the two sides AB, BH, are equal to the two sides DE, EF, each to each. But the angle ABH is equal (Hyp.) to the angle DEF. Therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the remaining angles of the one are equal to the remaining angles of the other, each to each, viz, those to which the equal sides are opposite. Therefore the angle BHA is equal to the angle EFD. But the angle EFD is equal (Hyp.) to the angle BCA. Therefore also the angle BHA is equal (Ax. 1) to the angle BCA; that is, the exterior angle BHA of the

triangle AHC is equal to its interior and opposite angle BCA; which is impossible (I. 16). Therefore BC is not unequal to EF; that is, BC is equal to EF. And AB is equal (Hyp.) to DE. Therefore the two sides AB, BC, are equal to the two sides DE, EF, each to each. And the angle ABC is equal (Hyp.) to the angle DEF. Therefore the base AC is equal (I. 4) to the base DF, and the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q. E. D.

PROP. XXVII. (THEOREM.)-If a straight line (EF) falling on two other straight lines (A B, CD) make the alternate angles (AEF, EFD) equal to each other, these two straight lines (AB, CD) are parallel. For, if AB be not parallel to CD, AB and CD being produced will meet either towards A and C, or towards B and D. Let AB, CD, be produced and meet towards B and D, in the point G. Then GEF is a triangle, and its exterior angle AEF is greater (I. 16) than its interior and opposite angle EFG. But the angle AEF is equal (Hyp.) to the angle EFG. Therefore the angle AEF is both greater than, and equal to the

A

E

B

C

angle EFG; which is impossible. Wherefore AB, CD, being produced, do not meet towards BD. In like manner, it may be proved, that they do not meet when produced towards A, C. But those straight lines in the same plane, which do not meet either way, though produced ever so far, are parallel (Def. 35) to one another. Therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXVIII. (THEOREM )-If a straight line (EF) falling upon two other straight lines (AB, CD), make the exterior angle (EGB) equal to the interior and opposite angle (GHD) upon the same side of the straight line; or make the two interior angles (BGH, GHD) upon the same side of it (EF), together equal to two right angles; these two straight lines (AB and CD) are parallel to one another.

Because the angle EGB is equal (Hyp.) to the angle GHD, and the angle EGB is equal (I. 15) to the angle AGH; therefore the angle AGH is equal (4x 1) to the angle GHD; and they are alternate angles. Therefore AB is parallel (I. 27) to CD.

Again, because the two angles BGH, GHD are together equal (Hyp) to two right angles; and the two angles AGH, BGH are also together equal (I. 13) to two right angles. Therefore the two angles AGH, BGH are equal (Ax. 1) to the two angles BGH, GHD. Take away from these equals, the common angle BGH.

A.

C

H

Therefore the remaining angle AGH is equal (Ax. 3) to the remaining angle GHD; and they are alternate angles. Therefore AB is parallel (I. 27) to CD. Wherefore, if a straight line, &c. Q. E. D.