Euclid's Elements of geometry, the first three books (the fourth, fifth, and sixth books) tr. from the Lat. To which is added, A compendium of algebra (A compendium of trigonometry).

Because the right lines AG and AC are equal (by Hypoth. and Constr.) the angles ACG and AGC will be equal (by Prop. 5): but BGC is greater than AGC, and therefore greater than ACG, and also greater than BCG.

Then in the triangle BGC, the angle BGC is greater than BCG, and therefore the side BC is greater than the side BG (by Prop. 19); but BG is equal to BD (by Constr. and Prop. 4), and therefore BC is greater than BD.

If two triangles (BAC and EFD) have two sides of the one equal to two sides of the other (BA to EF and AC to FD), but the remaining side (BC) greater than the remaining side (ED,) the angle A opposite the greater side will be greater than the angle (F) opposite the less.

The angle A is either equal to F, or less than it, or greater than

it.

It is not equal to it; for if it were, the side BC should be equal to the side ED (by Prop. 4), contrary to Hypothesis.

да

It is not less; for if it were, the side BC should be less than ED (by Prop. 19), contrary to Hypothesis. Therefore because the angle A is neither equal to the angle F, nor less than it, it will be greater.

If two triangles (BAC, DEF) have two angles of the one equal to two angles of the other (B to D and C to F); and a side of one equal to a side of the other, that is, either the sides which are between the equal angles (as BC to DF) or opposite to the equal angles (as BA to DE), the remaining sides and angle of the one are equal to the remaining angle and sides of the other.

First, let the side BC be equal to the side DF, then the side BA will be also equal to the side DE.

For they are not un

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equal, but if it be possible to be done, let one of them BA be greater than the other, and cut off the right line BG equal to the less DE, and draw CG. In the triangles GBC, EDF, the sides GB, BC are equal to the sides ED, DF (by Constr. and Hypoth.), and the angle B is equal to the angle D (by Hypoth.), therefore the angles GCB and F are equal (by Prop. 4); but the angle BCA is equal to the angle F (by Hypothesis), therefore BCG is equal to BCA (by Ax. 1), which is absurd; therefore neither of the sides BA and DE is greater than the other; BA and DE are therefore equal; but BC and DF are equal (by Hypoth.), and the angles B and D are equal (by Hypoth.), therefore the remaining side AC is equal to the remaining side EF, and also the remaining angle A to the remaining angle E (by Prop. 4.)

Again, let the equal sides BA and DE be taken, which are opposite to the equal angles C and F; and the sides BC and DF will be also equal.

For they are not unequal, but if it be possible to be done, let one of them BC be greater than the other, and cut off the right line BD equal to the less DF, and draw AD.

In the triangles ABD, EDF, the sides AB, BD are equal to the sides ED, DF (by Constr. and Hypoth.) and the angle B is equal to the angle D (by Hypoth.) therefore the angles ADB and F are equal (by Prop. 4), but the angle C is equal to the angle F (by Hypoth.), ADB is therefore equal to C (by Ax. 1); which cannot be possible (by Prop. 16); therefore neither of the sides BC and DF is greater than the other, BC and DF are therefore equal; BA and DE are equal (by Hypoth.),

and the angles B and D are equal (by Hypoth.), therefore the remaining side AC is equal to the remaining side EF, and the remaining angle A to the remaining angle E (by Prop. 4.)

SCHOL. It appears that the triangles themselves are also equal.

COR. 1.-In an Isosceles triangle ABC, a right line BD drawn from the vertex perpendicular to the base, bisects both the base and the vertical angle.

A D

For in the triangles ABD, CBD, the angles A and ADB are equal to the angles C and CDB (by Hypoth.), but the side BD, opposite to the equal angles A and C, is common, therefore the angles ABD, CBD are equal, and also the sides AD and DC (by Prop. 26); the vertical angle and the base are therefore bisected.

COR. 2.-It appears from Proposition 4th of this Book, that the right line bisecting the vertical angle of an isosceles triangle, also bisects the base, and is perpendicular to it; and from Prop. 8, that the right line drawn from the vertical angle bisecting the base, is perpendicular to it, and bisects the vertical angle.

If a right line (EF) cutting two right lines (AB and CD), make the alternate angles equal (AEF to EFD) these right lines will be parallel.

For suppose them not to be parallel, but meeting, if it be possible, in G, and the external angle AEF of the triangle EGF, is greater than the internal angle EFG (by c Prop. 16), but it is equal (by

Hypoth.), which is absurd; therefore the right lines. AB and CD do not meet towards B, D. It can be

similarly demonstrated, that they do not meet at the side A, C. Therefore, since they meet at neither side, they are parallel (by Def. 28.)

If a right line (EF) cutting two right lines (AB and CD,) make the external angle equal to the internal opposite angle on the same side of the line EF (EGA to GHC, or EGR to GHD); or make the internal angles at the same side of the line (AGH and CHG, or RGH and DHG) equal to two right angles, these right lines will be parallel.

First, let EGA and GHC be equal; and because EGA is equal to RGH (by Prop. A 15), GHC and RGH will be equal, and are alternate angles, the right lines AR and CD are therefore parallel (by Prop. c27).

It may be similarly demonstrated, if the angles EGR and GHD be equal.

Now let AGH and CHG together be equal to two right angles; and because GHD and CHG are also equal to two right angles (by Prop. 13), AGH and CHG together will be equal to GHD and CHG taken together (by Ax. 1), take away the common angle CHG and AGH will be equal to GHD, but they are alternate angles, therefore the lines AR and CD are parallel (by Prop. 27.)

It can be similarly demonstrated, if RGH and DHG be equal to two right angles.

A right line (EF) cutting parallel right lines, (AB and CD) makes the alternate angles equal (AGH to GHD, and CHG to HGB); and the external angle equal to the internal opposite angle EGA to GHC, and EGB to GHD); and also the internal angles at the same side of the right line (AGH and CHG, BGH and DHG) equal to two right angles.

First, the alternate angles AGH and GHD are equal. For if not, let one of them, AGH be greater than the other; and by adding BGH to both, AGH and BGH together will be greater c. than BGH and GHD; but AGH and BGH are equal to two right angles (by Prop. 13), therefore BGH and GHD are less than two right angles; and therefore the right lines AB and CD would meet at the side B,D (by Ax. 12), but they are parallel (by Hypoth.), and therefore cannot meet, which is absurd; therefore neither of the angles AGH and GHD is greater than the other; they are therefore equal. It can be similarly demonstrated, that BGH and GHC are equal.

Secondly, the external angle EGB is equal to the internal GHD. For EGB is equal to the angle AGH (by Prop. 15), and AGH is equal to the internal angle GHD (by part first), therefore EGB is equal to GHD. It can be similarly demonstrated, that EGA and GHC are equal.

Thirdly, the internal angles at the same side, BGH and GHD, are equal to two right angles. For since the alternate angles GHD and AGH are equal (by part 1st), by adding BGH to both, BGH and GHD will be equal to BGH and AGH, and therefore equal to two right angles (by Prop. 13.) It can be similarly demonstrated, that the angles AGH and GHC are equal to two right angles.

PROPOSITION XXX. THEOREM.

If two right lines (AB, CF) be parallel to the same right line (DN), they will be parallel to one another.

For let the right line GP cut them. The external angle GLB is equal to the internal LON (by Prop. 29), and the angle LON is similarly equal to the angle OPF, therefore GLB is equal to c—

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Euclid's Elements of geometry, the first three books (the fourth, fifth, and ...