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).

OPF (by Ax. 1), and the right lines AB and CF are therefore parallel (by Prop. 28.)

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

Draw the right line CF, cutting the given line AB. At the point C, and with the right line CF, construct an angle FCE, equal to the angle AFC (by Prop. 23), but at the opposite side Aof the right line CF; DE will be parallel to the given right line AB.

For the right line CF, cutting the lines DE and AB, makes the alternate angles ECF and AFC equal, and therefore these right lines are parallel (by Prop. 27.)

PROPOSITION XXXII. THEOREM.

In every triangle, if any side (AB) be produced, the external angle (FBC) is equal to the two internal remote angles (A and C). And the three internal angles are equal to two right angles.

Through B draw BE parallel to AC (by в Prop. 31). The angle FBE is equal to the internal remote angle A (by Prop. 29), and the angle EBC is equal to the alternate angle C (by Prop. 29), therefore the A whole angle FBC is equal to the two angles A and C.

The angle ABC with FBC is equal to two right angles (by Prop. 13), but FBC is equal to the two angles A and C (by part first), therefore the angle ABC with the two angles A and C, is equal to two right angles.

COR. 1.-If in a triangle, one angle be right, the remaining two together are equal to a right angle.

COR. 2.-If in two triangles, there be two angles in each respectively equal, the remaining angles will be also equal.

COR. 3.-In an isosceles right angled triangle, each angle at the base, is half a right angle.

COR. 4.-In an equilateral triangle, each angle is a third part of two right angles.

COR. 5.-Hence it is possible to trisect a right angle FAC. For assume any part of the leg AC, and construct an equilateral triangle CBA, whose angle CAB bisect with the right line AE; and because CAB is a third part of two right angles (by Cor. 4), it will be equal to two-thirds of one right angle, and therefore BAF is a third part of a right angle, and BAE, EAC, and FAB are therefore equal.

COR. 6.-All the angles together of any rectilineal figure ABCDE, are equal to twice as many right angles, deducting four, as the figure hath sides.

For assume a point F within the figure, and draw the right lines FA, FB, FC, FD and FE. There are as many triangles constructed as the figure has sides, and therefore all these angles will be equal to twice as many right angles as the figure has sides (by Prop. 32); from these take four right angles, for the angles at the point F (by Cor. 3 Prop. 13), and the remainder, namely the angles of the figure, will be equal to twice as many right angles, deducting four, as there are sides of the figure.

COR. 7.-All the external angles of any rectilineal figure, are equal to four right angles. For each external angle with the internal adjacent, are equal to two right angles (by Prop. 13); therefore all the

external with all the internal, are equal to twice as many right angles, as there are sides of the figure; but the internal with four right angles, are equal to twice as many right angles as there are sides of the figure (by Cor. 6); take away from both the internal angles, and the external angles will be equal to four right angles.

PROPOSITION XXXIII. THEOREM.

Right lines (AC and BD) joining the adjacent extremities of equal and parallel right lines (AB and CD) are themselves equal and parallel.

For draw the diagonal AD, and in triangles c CDA, BAD, the sides CD and BA are equal (by Hypoth.), but AD is common, and the angle CDA is equal to the alternate angle BAD (by Prop. 29), therefore the right lines AC and BD are equal (by Prop. 4); therefore the right line AD, cutting the right lines AC and BD, makes the alternate angles equal, and AC and BD are therefore parallel (by Prop. 27.)

The opposite sides (AB and CD, AC and BD) of a parallelogram (AD) are equal, and the opposite angles (A and D, C and B) are also equal, and it is bisected by a diagonal.

For in the triangles CDA, BAD, the angles CDA and BAD, CAD and BDA are equal, for they are the alternate angles (by Prop. 29), and the side AD between the equal angles is common to both triangles, therefore the sides A CD and CA are equal to AB and BD (by Prop. 26), and the triangle CDA is equal to the triangle BAD (by Sch. Prop. 26), and the angles ACD and ABD are also equal (by Prop. 26); and because ACD together with CAB is equal to two right angles (by Prop. 29), and ABD with CDB is also equal to two right angles, if the equals ACD and ABD be taken from both, the remaining angles CAB and CDB will be also equal.

COR. 1.-In any parallelogram if one angle be right, the remaining angles will be right angles.

For the adjacent angle is right, because with a right angle it is equal to two right angles, (by Prop. 29); and the opposite angles are right, since they are equal to these right angles (by Prop. 34.)

COR. 2.-If two parallelograms have an angle of the one equal to an angle of the other, the remaining angles will be also equal; for the angles which are opposite to these equal angles are equal to them (by Prop. 34), and therefore are equal to one another; but the adjacent angles are also equal to one another, because together with these equals, they are equal to two right angles (by Prop. 29.)

PROPOSITION XXXV. THEOREM. Parallelograms (BD and BF) on the same base and between the same parallels are equal.

Because the

right lines AD

and EF are equal to the same right line

BC (by Prop.34), they are equal to each other, therefore, by adding DE to both, as in figure 2, and by subtracting DE from both, as in figure 3, the right lines AE and DF are equal; also the right lines BA and BE are equal to the right lines CD and CF (Prop. 34), therefore the triangles BEA and CFD are equal (by Schol. Prop. 8); but subtract BEA from the quadrilateral BAFC, the remainder is the parallelogram BF; and subtract the triangle CFD from the same quadrilateral, the remainder is the parallelogram BD; therefore the parallelograms BD and BF are equal (by Ax. 3.)

Parallelograms (BD and EG) on equal bases, and between the same parallels, are equal.

For draw BF and CG, and because BC and FG are equal to the same EH (by Hypoth. and Prop. 34), they are equal to each other, but they are also parallel (by Hypoth.), therefore BF and CG joining them, are parallel (by

A D F G

C E H

Prop. 33), and BG is a parallelogram, therefore it is equal to both BD and EG (by Prop. 35), and therefore BD and EG are equal to one another. (by Ax. 1.)

Triangles (BAC and BDC) on the same base, and between the same parallels, are equal.

For draw through the point C, the right lines CF and CE, parallel to BA and BD (by Prop.31), and the parallelograms BAEC, BDFC will be equal (by Prop. 35), but the triangles BAC and BDC are halves of them (by Prop. 34), and therefore are also equal (by Ax. 7.)

PROPOSITION XXXVIII.

A ED

B C

THEOREM.

Triangles (BAC and HDE) on equal bases, and between the same parallels, are equal.

A FD G

For draw through the points C and E the right lines CF and EG, parallel to the right lines BA and HD (by Prop. 31); and the parallelograms BAFC, HDGE, will be equal (by Prop. 36), but the triangles BAC and HDE are halves of them, (by Prop. 34), and therefore are also equal (by Ax. 7.)

B C H E

PROPOSITION XXXIX. THEOREM.

Equal triangles (BAC and BDC) on the same base, and on the same side of it, are between the same parallels.

For if AD be not parallel to BC, draw A through the point A the right line AF parallel to BC, cutting the side BD of the triangle BDC in the point E, different from the vertex, and draw CE.

Because AF and BC are parallel, the triangle BEC is equal to the triangle BAC (by Prop. 37), but BDC

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