which is absurd [16. 1]; therefore AB, DE are not unequal, they are therefore equal; whence, AC being equal to DF, and the angle A to the angle D [Hyp.], BC is equal EF, and the angle ACB to the angle F [4.1].

Cor.-A right line [CD], drawn from the vertex [C] of an isosceles triangle [ACB], bisecting the base [AB], is perpendicular to it; and, if it be perpendicular to the base, it bisects it.

Part 2.-Let CD be perpendicular to AB, it bisects it. In the triangles ACD, BCD, the angles A and B are equal (5. 1), as are also the angles ADC, BDC [Hyp. and Def. 20], and CD opposite the equal angles A, B is common, therefore AD is equal to DB (26. 1).

Schol. Only two equal right lines (CA; CB) can be drawn from the same point (C) to any right line (AB); since any two right lines, drawn from C to AB, on the same side of the perpendicular CD, are unequal [Cor. 2. 19. 1]. And these equal Fight lines form equal angles with that to which they are drawn (5 1); and two right lines (CA, CB), drawn to any right line (AB), from a point (C) without it, and making equal angles (CAB, CBA) with it, are equal (6. 1).

PROP. XXVII. THEOR.

If a right line (EF), meeting two other right lines (AB, CD), make the alternate angles (AEF, EFD) equal; the right lines (AB, CD) so met, are parallel.

For, if AB and CD be not parallel, they may be so produced, as to meet either towards B, D or A, C [Def. 34.]; let them, if possible, being produced towards B,D, meet, as in G, and the external angle AEF, of the triangle EFG, is greater than the internal remote one EFG (16. 1), but it is also equal to it [Hyp.], which is absurd; therefore AB, CD do not meet towards B, D. In like manner it may be shewn, that they do not meet towards A, C: since therefore they do not meet towards either part, they are parallel.

PROP. XXVIII. THEOR.

If a right line (EF), cutting two other right lines (AB, CD), make an external angle (EGB), equal to the internal remote on the same side (GHD), or two internals on the same side (BGH, GHD) equal to two right angles; the right lines (AB, CD) so cut, are parallel.

First-Let EGB be equal to GHD; and because AGH is equal to EGB (15. 1), AGH is equal to GHD [Ax. 1], and they are alternate angles, therefore AB and CD are parallel (27. 1).

Let now the angles BGH, GHD be equal to two right angles; and because AGH, BGH are also equal to two right angles (13. 1), the angles BGH, GHD, together, are equal to AGH, BGH together; take away from each the common angle BGH, and the angle AGH is equal to GHD, and they are alternate angles, therefore AB and CD are parallel (27. 1).

PROP. XXIX. THEOR.

A right line (EF, see fig. to the preced. prop.), cutting two parallel right lines (AB, CD), makes the alternate angles equal, (AGH to GHD, and BGH to GHC); any external angle (as EGB), to the internal remote on the same side (GHD); and any two interior angles on the same side (as BGH, GHD, together equal to two right angles.

First. The alternate angles, as AGH, GHD, are equal. For if not, let one of them, as AGH, be the greater; and the angles AGH, HGB being together equal to two right angles (13. 1), or, which is equal (13. 1. and Theor. at 11. 1), to CHG, GHD; taking from them the unequals AGH, GHD, the remainder CHG is greater than the remainder BGH [Ax. 5]; whence, AGH being greater than GHD [Hyp.], AGH, GHC together are greater than BGH, GHD together, therefore AB, CD may be so produced towards the part B, D, as to meet [Ax. 12], which is absurd [Hyp. and Def. 34]; therefore the angles AGH, GHD are not unequal, they are therefore equal. In like manner the angles BGH, GHC may be proved equal.

Secondly. An exterior angle, as EGB, is equal to the interior remote on the same side GHD. For the angle EGB is equal to AGH [15. 1], and AGH is equal to the alternate GHD [part 1. of this], therefore EGB is equal to GHD [Ax. 1].

Thirdly. Two interior angles on the same side, as BGH, GHD, are equal to two right angles. For the alternate angles AGH, GHD are equal [part 1, of this], adding to each BGH, the angles BGH, GHD together, are equal to AGH, BGH together [Ax. 2.]; but AGH, BGH together are equal to two right angles [13. 1], therefore BGH, GHD together are equal to two right angles.

Theorem. If a right line [EF], meeting two other right lines [AB, CD], make the interior angles on one side of it [BGH, GHD less than two right angles; these right lines may be so produced towards the part [B, D], on which the angles are less than two right angles, as to meet.

For, because the right line HG falls on AB, the angles AGH, HGB are together equal to two right angles [13. 1]; for the same reason, the angles CHG, GHD are together equal to two

right angles; therefore the four angles AGH, HGB, CHG, GHD are together equal to four right angles; but the angles BGH, GHD are together less than two right angles [Hyp.], therefore the angles AGH, GHC are together greater than two right angles, and therefore greater than the angles BGH, GHD together, and of course the right lines AB, CD may be so produced towards the part B, D, as to meet [Ax. 12].

PROP. XXX. THEOR.

Two right lines (AB, CD), which are parallel to the same right line (EF), are parallel to each other.

Draw the right line GHK, cutting AB EF and CD in G, H and K. And, because AB, EF are parallel, and GK cuts them, the angle AGH is equal to the alternate GHF [29. 1]; and, because GK cuts the parallels EF, CD, the angle GHF is equal to the internal remote HKD [29. 1[; whence, the angles AGH, HKD, being each equal to GHF, are equal to each other [Ax. 1]; and therefore AB and CD are parallel [27. 1].

PROP. XXXI. THEOR.

To a given right line (BC), through a given point (A) without it, to draw a parallel right line.

To any point D in BC, join AD, and at the point A, with the right E line AD, make the angle DAE equal to ADC [23. 1], on contrary sides of the right line AD; the right line EAF is parallel to BC.

B

A

-F

For the right line AD, meeting the right lines EF, BC, makes the alternate angles EAD, ADC equal; therefore EF is parallel to BC [27. 1].

PROP. XXXII. THEOR.

If any side (BC) of a triangle (ABC) be produced, the exterior angle (ACD) is equal to the two interior remote ones (A and B). And the three interior angles (A, B and ACB), of every triangle (ABC), are equal to two right angles.

A

B

Through C draw CE parallel to BA [31. 1]. Because AC meets the parallels BA, CE, the alternate angles ACE, BAC are equal [29. 1]; and because BD meets the same parallels, the external angle ECD is equal to the interior remote on the same side ABC [29. 1]; therefore the angles ACE, ECD together, or the angle ACD, is equal to the two angles A and B together [Ax. 2],

To each of these equals add the angle ACB, and the two angles ACD and ACB are together equal to the three angles A, B and ACB [Ax. 2]; but the angles ACD and ACB are together equal to two right angles [13. 1], therefore the angles A, B and ACB are equal to two right angles.

Cor. 1. All the interior angles (ABC, BCD, CDE, DEA and EAB) of any rectilineal figure (ABCDE), are equal to twice as many right angles, except four, as the figure has sides.

Take any point F within the figure, and draw FA, FB, FC, FD, FE; there are

D

formed as many triangles as the figure has sides, all the angles of which are equal to twice as many right angles as the figure has sides [by this prop.]; but of these all the angles.

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