PROPOSITION IV.-PROBLEM.

To make a triangle AEB, whose three sides shall be equal to the three given straight lines, AB, C, and D.

From the centre A, with a radius equal to C, describe the circle EFH, (Post. 3), and from the centre B, with a radius equal to D, describe a circle EGH; and from the point E, where the circles

cut each other, draw the Is AE and EB, (Post. 1), then AEB will be the triangle required.

Because AE is the radius of the circle EFH, and it was described with a radius =C, .. AE is =C, and because BE is the radius of the circle EGH, and it was described with a radius =D, .. EB is =D. Hence the ▲ AEB has its three sides to the three |s AB, C, and D.

Cor. 1. If the Is C and D were equal, the triangle would be isosceles.

Cor. 2. If the given lines, AB, C, and D, were all equal, the triangle would be equilateral.

Cor. 3. If C and D were together less than AB, the circles would not intersect, and the construction would be impossible; hence any two sides of a triangle are together greater than the third, which is established in a different manner in (Prop. 13.)

PROPOSITION V.-THEOREM.

If two triangles, ABC and DEF, have two sides, AC and CB, and the contained angle ACB in one, respectively equal to two sides, DF and FE, and the contained angle DFE in the other, the triangles are equal in every respect. For conceive the point C to be laid on the point F, and the CA on the FD, then,

F

ДД

BD

E

these lines are =, the point A will coincide with the point D. And since CA coincides with FD, and the LC is the LF, the line CB will fall on the line FE; and CB and FE are, the point B will coincide with the point E. Hence since

the points A and B coincide with the points D and E, the line AB will coincide with the line DE, and the A ABC will coincide with the ADEF; .. the two As are equal, and have all the parts of the one equal to the corresponding parts of the other, namely the side AB = the side DE, the LA≈ the LD, and the LB the LE. Q. E. D.

PROPOSITION VI.-THEOREM.

In any two triangles, ABC, DEF, if two angles, A and B in the one, be respectively equal to two angles, D and E in the other, and the sides AB and DE, which lie between these equal angles be also equal, the triangles are equal in all respects.

A4

BD

F

E

For conceive the point A to be laid on the point D, and the side AB on the side DE, then, ... these lines are equal, the point B will coincide with the point E. And as AB and DE coincide, and the LA is the LD, the side AC will coincide with DF, and for a like reason BC will coincide with EF. ..since AC falls on DF, and BC on EF, the point C must coincide with the point F; and ..the two As are in all respects equal, having the other sides, AC, BC, the two, DF, EF, and the remaining LC the remaining LF.

In an isosceles triangle, ABC, the angles A and B, opposite the equal sides BC and AC, are equal.

For, conceive the LC to be bisected by the line CD, then the two As ACD and BCD have AC-BC, and CD common to both, and the LACD = E

..

Q. E. D.

the LBCD; they are equal in every respect, (Prop. 5), and have the A= the LB, they are opposite to the common side CD. Q. E. D. Cor. 1. If the equal sides AC and BC be produced to E and F, the angles EAB and ABF on the other side of the base will also be equal.

For, the two Ls CAB, BAE, are together two rLs, (Prop. 1), and the two Ls CBA and ABF are also together to two Ls; .. the two Ls CAB, BAE, are the two Ls CBA, ABF, (Ax. 1); and it was proved that the

Ls CAB and CBA are equal; .*. the angles EAB and ABF are also equal, (Ax. 3.).

Cor. 2. The line that bisects the vertical angle of an isosceles triangle also bisects the base, and cuts it at right angles.

For, the As ACD and BCD were shown to be equal in every respect ; .. AD is = BD, and LADC is = BDC, and they are adjacent Ls; hence (Def. 9) each of them is a right angle.

Cor. 3. Every equilateral triangle is also equiangular.

PROPOSITION VIII. THEOREM.

If two angles, CAB and CBA, of a triangle be equal, the sides, CB and CA, opposite them will also be equal. For, if AC be not CB, let AC

be CB, and let AD be the part of
AC that is BC, join BD, (Post. 1);
then the As ADB, CBA, have AD
=CB, and AB common; and the
LDAB contained by the two sides of
the one is the LCBA contained by
the two sides of the other; .. (Prop. 5) AL
the ABD is the AABC, the

D

B

BC,

less the greater, which is impossible; .. AC is not and in the same manner it may be shown it is not less; hence AC is = BC.

Q. E. D.

Cor. Every equiangular triangle is also equilateral.

PROPOSITION IX.-THEOREM.

If two triangles, ABC and DEF, have the three sides of the one respectively equal to the three sides of the other, the triangles shall be equal in all respects, and have those angles equal, that are opposite to equal sides.

Let ABC and DEF be two

As, having AC=DF, CB=FE, and AB-DE, and let AB and DE be the sides which are not A

less than any of the others. Conceive the side DE to be applied to the side AB, so that the point D may coincide with A, and the

C

G

F

B D

E

line DE with AB, then the point E will coincide with B, ·.· DE is = AB; but let the vertex F fall in the opposite direction, from C as at G, join GC; then. CB is = FE,

is

and FE is = BG, being the same line in a different position, CB is =BG, (Ax. 1); .. the LBCG is LBGC, (Prop. 7); again, ... AC is=DF and DF is AG, .. AC AG, and hence the LACG is the LAGC; .. also the whole LACB is the LAGB, (Ax. 2); but LAGB is the LDFE in a different position; .. the LACB is = the LDFE; and since AC, and CB, and the LC, are respectively = DF and FE and the LF, (Prop. 5), the LA is the LD, and the LB is the LE. Hence the angles are equal that are opposite to the equal sides.

Q. E. D.

Cor. 1. The areas of the triangles are also equal. Cor. 2. To make at the point F, in the straight line DF, an angle equal to ACB. Construct a ADFE, (Prop. 4.) having its three sides equal to the three straight lines AC, CB, BA, namely FD=AC, FE=CB, and ED=AB; the LDFE will be the LACB by the proposition.

PROPOSITION X.-THEOREM.

If a side BC of a triangle ABC be produced to D, the exterior angle ACD will be greater than either of the interior opposite angles CAB or ABC.

Conceive AC to be bisected in E, and BE joined, and the line BE produced to F, so that EF may be =BE and join FC, and produce AC to G; then the As AEB and CEF have AE and EB, and the contained LAEB in the one = CE and EF, and the contained

the

LCEF in the other; . (Prop. 5) the LEAB is LECF; but LACD is greater than LECF; .. (Ax. 13) the LACD is greater than the LBAC.

If the side BC were bisected, and a similar construction made below the base, it might be shown in the same manner, that the LBCG, which is the LACD, (Prop. 3), is the LABC; .. the LACD is either of the LSCAB

or ABC. Q. E. D. Cor. Any two angles of a triangle are together less than two right angles.

For the LACD is the LBAC, and if the LACB be added to each, the two LSACD and ACB (that is, two Ls, Prop. 1.) are the LSBAC and ACB; and the same might be shown of any other two angles.

G

PROPOSITION XI.-THEOREM.

The greater side of every triangle has the greater angle opposite to it.

be

If the side AC of the AABC

the side AB, the LABC

will be the LACB.

For make AD-AB, and join

BD, then the LADB is

the

LABD(Prop.7); but the LADB B

is the LACB (Prop. 10); ..

D

the LABD is the L ACB (Ax. 12), still more then is the whole LABC the LACB.

Q. E. D. Cor. The greatest side of any triangle has the greatest angle opposite to it.

PROPOSITION XII.-THEOREM.

If the angle ABC of the triangle ABC be greater than the angle ACB, the side AC opposite the greater angle will be greater than the side AB opposite the less. B

C

Or the greater angle of every triangle has the greater side opposite to it.

less;

the

For if AC be not AB, it must either be it or AC is not = AB, for then the LB would be LC (Prop. 7), which it is not; neither is AC

AB, for then the LB would be the LC, which it is not (Prop. 11), .. AC is > AB.

Cor. The greatest angle of every triangle has the greatest side opposite to it.

PROPOSITION XIII.-THEOREM.

In any triangle ABC, the sum of any two of its sides, as AB and AC, is greater than the remaining side BC.

Produce AB to D, so that AD may be = AC, and join DC; then. AD is = AC(by Const.), the L ACD is the LADC (Prop. 7), .. in the DBC the LBCD is

=

B

the LBDC, hence the side BD is the side BC (Prop. 12); but BD is = AB and AC, ·.· AD is = AC, .. BA and AC are together BC (Ax. 12.)

Q. E. D. Cor. The difference of two sides of a triangle is less than the third side.

For, since AB and AC are⇒BC, if AC be taken from each there remains AB the difference of BC and AC, (Ax. 5.)