about the point F are equal to four right angles [Cor. 2. 15. 1]; therefore the remaining angles, which constitute the interior angles of the figure, are equal to twice as many right angles, except four, as the figure has sides.

Cor. 2.-All the exterior angles, of any rectilineal figure (AC), are together equal to four right angles.

For every exterior, as ABD and its adjacent interior ABC, are together equal to two right angles [13. 1]; therefore all the exterior and interior angles of the figure are together equal to twice as many right angles as the figure has sides; but the interior angles are equal to twice as many right angles, except four, as the figure has sides [by the preceding Cor.], therefore the exterior angles are equal to these four.

Cor. 3.-If from the vertex (C, see fig. 1, 2 and 3), of an isosceles triangle (ABC), a right line (CD) be drawn without the triangle, equal to one of its equal sides (AC or CB), the angle (ADB) formed at its other extreme (D), by right lines (DA, DB) drawn to the extremes (A, B) of the base, is equal to half the vertical angle (ACB).

First, let one of the right lines, drawn from D to the extremes of the base, pass by the vertex C, as in fig. 1; and, since the triangles CBD is isosceles, the angles CBD, CDB are equal [5. 1], whence, the external angle ACB of the tri

angle BCD, being equal to the two internal remote angles CBD, CDB [by this prop.], is double to ADB.

Secondly, let the point C be within the triangle ADB, as in fig. 2. Draw DC, which produce beyond C, as to E; and, by case 1, the angle ACE is equal to the double of ADC, and BCE to the double of BDC, and therefore, the whole ACB to the double of the whole ADB [Ax. 2].

Thirdly, let the point C be without the triangle ADB, as in fig. 3. Draw DC, which produce beyond C, as to E; the angle ECB is [by case 1,] equal to the double of EDB, and ECA to the double of EDA; therefore, taking the latter from the former, ACB is equal so the double of ADB [Ax. 3].

Cor. 4.-If the angle (ADB, see fig. 1. 2 and 3,] formed at any point (D), above the base (AB), of an isosceles triangle (ACB), by right lines (DA, DB) drawn to the extremes of the base. be equal to half the vertical angle (ACB); the right line (CD), joining that point to the vertex of the triangle, is equal to one of the equal sides of the triangle (AC or CB).

First, let CD be in the same right line with one of the equal sides AC, as in fig. 1 above; then, since ACB is equal to CBD and CDB [by this prop.], and CDB equal to the half of ACB [Hyp.], CBD is also the half of ACB; therefore the angles CBD, CDB are equal [Ax. 7], and therefore CD is equal to CB [6. 1].

Secondly, let the angle ACB be included within the angle ADB, as in fig. 2 above; CD is in this case also equal to CB; for, if not, it is either greater or less than it; let CD, if possible, be greater than CB, and take thereon CF equal to CB [3. 1], and join FA, FB; the angle AFB is equal to half the angle ACB [by preceding Cor.], but ADB is equal to half the angle ACB [Hyp.], therefore the angles AFE, ADB are equal [Ax. 7], which is absurd [21. 1], therefore CD is not greater than CB; in like manner it may be shewn, that CD is not less than CB; it is therefore equal to it.

Thirdly, let C be without the triangle ADB, as in fig. 3 above; CD is in this case also equal to CB; for, if not, it is either greater or less than it; and first, let CD, if possible, be greater than CB, and on it take CF equal to CB; from the centre C, at the distance CF, describe an arch of a circle FG [Post. 3], which, (because CF and CB are equal), being continued, would pass through B, whence, the points F, B of the circumference, being on contrary sides of AD, that circumfe

rence must meet AD between these points; let it meet AD in G, and join FA, CG, GB; and because CG is equal to CB [Def. 10], the angle AGB is half of ACB [by last Cor.], but ADB is also half of ACE [Hyp.], therefore the angles AGB, ADB are equal (Ax. 7, the external angle AGB of the triangle GBD to the internal remote GDB, which is absurd (16. 1); therefore CD is not greater than CB: Let now CD, if possible, be less than CB, and, on CD produced, take CH equal to CB; from the centre C, at the distance CH, describe an arch of a circle, meeting AD produced in K, and join KB; and because a right line joining the points C, K is equal to CH (Def. 10); or its equal (Constr.) CB, the angle AKB is half the angle ACB [by preced. Cor.], but the angle ADB is also half of ACB (Hyp.), therefore the angles ADB; AKB are equal (Ax. 7), the external angle ADB of the triangle DBK to the internal remote DKB, which is absurd (16. 1); therefore CD is not less than CB; and it is above shewn not to be greater than ('B; since therefore CD is neither greater nor less than CB, it is equal to it.

PROP. XXXIII. PROB.

Right lines (AC, BD), joining the adjacent extremes of equal and parallel right lines (AB, CD), are themselves equal and par

Join CB, and in the triangles ABC, BCD, the sides AB; CD, are equal [Hyp.], BC common, and, because A and CD are parallel [Hyp.], the alternate angles ABC, BCD are equal [29. 1], therefore [by 4. 1,] AC is equal to BD, and the angle ACB to CBD; but these are alternate angles, formed by the right line BC meeting the right lines AC, BD; therefore AC and BD are parallel [27. 1].

PROP. XXXIV. THEOR.

The opposite sides (AB, CD, and AC, BD), and opposite angles (A, D, and ABD, ACD), of a parallelogram (AD), are equal, and a diagonal, or right line, joining two of its opposite angles (as CB), bisects it.

In the triangles ABC, BCD, the angles ABC, BCD are equal, as also the angles ACB, CBD, being alternate angles, formed by BC meeting the parallels AB, CD, and AC, BD [29. 1]; but the side CB, between the equal angles, is common to both these triangles, therefore AB is equal to CD, AC to BD, and the angle A to D [26. 1]; and, because AB, CD are parallel, the angles A and ACD are equal to two right angles [29. 1], or, which is equal [29. 1 and Theor. at 11, 1], to the angles D and ABD; taking from each, the equal angles A, D, the remaining angles ACD, ABD are equal [Ax. 3].

And, since AB, BC and the included angle ABC, are severally equal to CD, CB and the included angle BCD, the triangles ABC, BCD are equal [4. 1]; therefore CB bisects the parallelogram AD.

Cor. 1.—If one angle of a parallelogram be right, the rest are right.

For either adjacent angle is right, because it and a right angle are equal to two right angles [29. 1]; and the opposite angles are right, because equal to these right angles [by this prop.]

Cor. 2.-Two parallelograms, which have one angle of one equal to one angle of another, are mutually equiangular.

For the angles, which are opposite these equal angles, are equal to them 34. 1], and therefore to each other; and the an

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gles which are adjacent to them, because with them they are equal to two right angles [29. 1], are also equal.

Cor. 3-Two parallelograms, which are mutually equiangular and equilateral, are equal.

For, drawing a diagonal in each, subtending equal angles, either of the two triangles into which one of the parrallelograms is divided, is equal to either of those into which the other is divided [4. 1], therefore the parallelograms, which are double to these triangles [34. 1], are equal [Ax. 6].

Cor. 4.-Every quadrilateral figure [ABDC], whose opposite sides are equal, [AB to CD, and AC to BD,] is a parallelogram.

Join CB, and, in the triangles ABC, DCB, the sides CA, AB are severally equal to BD, DC [Hyp.], and BC is common, therefore the angles A, D are equal [8. 1]; whence also, the angle ABC is equal to BCD, and ACB to CBD [4. 1], therefore AB is parallel to CD, and AC to BD [27. 1], and ABDC is a parallelogram [Def. 35].

Cor. 5.-Every quadrilateral figure [ABDC], whose opposite angles are equal, [A to D, and ABD to ACD,] is a parallelogram.

Because the angle A is equal to D, and ABD to ACD [Hyp.], the four angles A, ABD, D and ACD are double to the two angles A and ABD; but the four angles A, ABD, D and ACD, of the quadrilateral figure ABCD, are equal to four right angles [Cor. 1. 32. 1]; therefore the angles A and ABD are equal to two right angles, and therefore AC and BD are parallel [28. 1]. In like manner it may be proved, that AB and CD are parallel; of course ABDC is a parallelogram [Def. 35].

Cor. 6. Two finite right lines [AB, AC], meeting in an angle, being given, to make a parallelogram, whereof they are sides.

Through B draw BD parallel to AC, and through C, CD parallel to AB [31. 1]; and, having joined CB, because the two angles ABC, ACB of the triangle ABC, are less than two 'right angles [17. 1], the two angles BCD, CBD, being alternates to them, and therefore severally equal to them [29. 1], are also less than two right angles; therefore BD, CD may be so produced as to meet [Theor. at 29. 1]; let them meet, as in D; the quadrilateral ABDC is a parallelogram, whereof the given right lines AB, AC are sides [Def. 35].