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L: 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 7CB, 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 A ABD is = the ABC, the less = the greater, which is impossible ; -, AC is not >BC, and in the same manner it may be shown it is not less; hence AC is

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 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,

= BC.

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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 is =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 AĆ, 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 bisec

A ted 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 LCEF in the other; :: (Prop. 5) the LEAB is = the 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 z the LABC; :. the LACD is 7 either of the ZsCAB or ABC.

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

For the LĂCD is 7 the LBAC, and if the LACB be added to each, the two LSACD and ACB (that is, two 1* Ls, Prop. 1.) are 7 the LsBAC and ACB; and the same might be shown of any other two angles.

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PROPOSITION XI.--THEOREM. The greater side of every triangle has the greater angle opposite to it.

If the side AC of the A ABC be > the side AB, the LABC will be → the LACB.

For make AD-AB, and join BD, then the LADB is = the L ABD(Prop.7); but the LADB B is 7 the LACB (Prop. 10); :. the LABD is 7 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.

For if AC be not > AB, it must either be less; AC is not = AB, for then the LB would be = the LC (Prop. 7), which it is not; neither is AC < AB, for then the Z B would be < the LC, which it is not (Prop. 11), .. AC is 7 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 L ADC (Prop. 7), ., in the ADBC the LBCD is

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 ACare > BC, if AC be taken from each there remains AB 7 the difference of BC and AC, (As. 5.)

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PROPOSITION XIV.-THEOREM. If two triangles, ABC, DEF, have two sides, AB, BC, of the one respectively equal to, DE and EF, two sides of the other, but the angle ABC, included by the two sides of the one, greater A than the angle DEF, included by the corresponding sides of the other; then the side AC is greater than the side DF.

Let ABG be the part of the LABC, which is = DEF, and let BG be = EF or BC. Then the As ABG, DEF, are equal in all respects, (Prop.5), and have the side AG= DF. And as BC and BG are =, the L BGCis = the LBCG (Prop. 7); but the LBCG is the ACG,.. also the LBGC is 7 the LACG, (Ax. 12); much more then is the LAGC 7 the LACG, and hence (Prop. 12), the side AC is z AG, and .. also 7 its equal DF (Ax. 13).

Q. E. D. PROPOSITION XV.-THEOREM. If two triangles, ABC, DEF, have the two sides AC, CB of the one respectively equal to two sides DF, FÈ of the other, but the remaining side AB of the one greater than the remaining side DE of the A other; the angle ACB will be greater than the angle DFE.

For, if the LACB be not > the ZDFE, it must either be equal to it or less; the LACB is not = DFE, for then (Prop. 5), the base AB would be DE, which it is not ; neither is the LACB < the ZDFE, for then (Prop. 14), the base AB would be the base DE, which it is not, :, the LACB is the ZDFE.

Q. E. D. PROPOSITION XVI.--THEOREM. If a straight line, fall upon

two other straight lines,

A. AB, CD, and make the alternate angles AEF, EFD, equal to one another, the straight lines AB and CD are parallel.

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For, if AB and CD be not ||, they will meet when produced, either towards B, D, or A, C; suppose that they meet towards B, D, in the point G, then EGF is a A, and its exterior LAEF is 7 the ZEFG (Prop. 10); but ZAEF is = _EFG, .. AB and CD do not meet towards B, D, and in the same manner it may be shown that they do not meet towards A, C, .. (Def.:37), AB is parallel to CD.

Q. E. D. Cor. 1. If the exterior angle HEB be equal to the interior angle EFD, AB is parallel to CD.

For, since LHEB is = LEFD, and also to LAEF, (Prop. 3), .. LAEF is = _EFD, and they are alternate angles; therefore (Prop. 16), AB is parallel to CD.

Cor. 2. If the two angles BEF and EFD be together equal to two right angles, AB and CD are parallel.

For, since the two Ls BEF and EFD are = two ni Ls, by (Sup.), and the two Ls BEF and AEF are together = two re _s (Prop. 1), the two Ls BEF and EFD are = to the two Ls BÈF and AEF, and taking away the common angle BEF, there remains the LAEF = the ZEFD, and they are alternate angles; hence AB is parallel to CD, (Prop. 16.)

PROPOSITION XVII.-THEOREM. If two parallel straight lines, AB and CD, be cut by another line EF, in the points G and H, the alternate angles AGH and GHD will be equal, c the exterior angle EGB will be equal

F to the interior opposite angle GID; and the two interior angles BGH and GHD on the same side of the line will be together equal to two right angles.

If the LAGH he not = the LGHD, let LG be drawn, making the LGH = the LGHD, and produce LG to M.

::: the LGH is =.the LGHD, and they are alternate Ls, :: LM is || to CD, (Prop. 16). But AB is given || to CD.

.. through the same point G there are drawn two straight lines LM and AB, || to the same straight line CD, which is impossible. (Ax. 16.) Hence the LAGH is not unequal to the LGHD, that is, it is equal to it.

Then, since the LAGH is = the ZGHD, and also to the EGB (Prop. 3), the LEGB is = the LGHD, exterior is equal to the interior opposite angle on the same side of the line.

Again, since the ZEGB is = the LGHD, to each of these equals add the ZBGH, then will the two Ls EGB,

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