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PROPOSITION XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.

Let ABC be a triangle, and let one of its sides BCbe produced to D. Then the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC:

and the three interior angles ABC, BCA, CAB shall be equal to two right angles.

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Through the point C draw CE parallel to the side BA. (I. 31.) Then because CE is parallel to BA, and AC meets them, therefore the angle ACE is equal to the alternate angle BAC. (1. 29.) Again, because CE is parallel to AB, and BD falls upon them, therefore the exterior angle ECD is equal to the interior and opposite angle ABC; (1. 29.)

but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC. (ax. 2.)

Again, because the angle ACD is equal to the two angles ABC, BAC, to each of these equals add the angle ACB,

therefore the angles ACD and ACB are equal to the three angles ABC, BAC, and ACB. (ax. 2.)

but the angles ACD, ACB are equal to two right angles, (I. 13.) therefore also the angles ABC, BAC, ACB are equal to two right angles. (ax. 1.)

Wherefore, if a side of any triangle be produced, &c. Q. E.D.

COR. 1. All the interior angles of any rectilineal figure together with four right angles, are equal to twice as many right angles as the figure has sides.

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For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles.

Then, because the three interior angles of a triangle are equal to two right angles, and there are as many triangles as the figure has sides, therefore all the angles of these triangles are equal to twice as many right angles as the figure has sides;

but the same angles of these triangles are equal to the interior angles of the figure together with the angles at the point F:

and the angles at the point F, which is the common vertex of all the triangles, are equal to four right angles, (1. 15. Cor. 2.) therefore the same angles of these triangles are equal to the angles of the figure together with four right angles;

but it has been proved that the angles of the triangles are equal to twice as many right angles as the figure has sides;

therefore all the angles of the figure together with four right angles,

are equal to twice as many right angles as the figure has sides. COR. 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles.

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Since every interior angle ABC with its adjacent exterior angle ABD, is equal to two right angles, (1. 13.)

therefore all the interior angles, together with all the exterior angles, are equal to twice as many right angles as the figure has sides;

but it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides;

therefore all the interior angles together with all the exterior angles, are equal to all the interior angles and four right angles, (ax. 1.) take from these equals all the interior angles, therefore all the exterior angles of the figure are equal to four right angles. (ax. 3.)

PROPOSITION XXXIII. THEOREM.

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and parallel straight lines,

and joined towards the same parts by the straight lines AC, BD. Then AC, BD shall be equal and parallel.

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Then because AB is parallel to CD, and BC meets them, therefore the angle ABC is equal to the alternate angle BCD; (1. 29.) and because AB is equal to CD, and BC common to the two triangles ABC, DCB; the two sides AB, BC, are equal to the two DC, CB, each to each, and the angle ABC was proved to be equal to the angle BCD: therefore the base AC is equal to the base BD, (I. 4.) and the triangle ABC to the triangle BCD,

and the other angles to the other angles, each to each, to which the equal sides are opposite;

therefore the angle ACB is equal to the angle CBD.

And because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another; therefore AC is parallel to BD; (1. 27.)

and AC was shewn to be equal to BD. Therefore, straight lines which, &c. Q. E.D.

PROPOSITION XXXIV. THEOREM.

The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it into two equal parts.

Let ACDB be a parallelogram, of which BC is a diameter. Then the opposite sides and angles of the figure shall be equal to one another; and the diameter BC shall bisect it.

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Because AB is parallel to CD, and BC meets them,

therefore the angle ABC is equal to the alternate angle BCD. (1. 29. And because AC is parallel to BD, and BC meets them, therefore the angle ACB is equal to the alternate angle CBD. (1. 29.) Hence in the two triangles ABC, CBD,

because the two angles ABC, BCA in the one, are equal to the two angles BCD, CBD in the other, each to each;

and one side BC, which is adjacent to their equal angles, common to the two triangles;

therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, (1. 26.)

namely, the side AB to the side CD, and AC to BD, and the angle BAC to the angle BDC.

And because the angle ABC is equal to the angle BCD,
and the angle CBD to the angle ACB,

therefore the whole angle ABD is equal to the whole angle ACD;
(ax. 2.)

and the angle BAC has been shewn to be equal to BDC; therefore the opposite sides and angles of a parallelogram are equal to one another.

Also the diameter BC bisects it.

For since AB is equal to CD, and BC common, the two sides AB, BC, are equal to the two DC, CB, each to each,

and the angle ABC has been proved to be equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD; (1.4.) and the diameter BC divides the parallelogram ACDB into two equal parts.

Q.E.D.

PROPOSITION XXXV. THEOREM.

Parallelograms upon the same base, and between the same parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC.

Then the parallelogram ABCD shall be equal to the parallelogram EBCF.

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If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D;

then it is plain that each of the parallelograms is double of the triangle BDC; (1.34.)

and therefore the parallelogram ABCD is equal to the parallelogram DBCF. (ax. 6.)

But if the sides AD, EF, opposite to the base BC, be not terminated in the same point;

Then, because ABCD is a parallelogram,
therefore AD is equal to BC; (1. 34.)

and for a similar reason, EF is equal to BC;
wherefore AD is equal to EF; (ax. 1.)

and DE is common;

therefore the whole, or the remainder AE, is equal to the whole, or the remainder DF; (ax. 2 or 3.)

and AB is equal to DC; (1. 24.)

hence in the triangles EAB, FDC,

because FD is equal to EA, and DC to AB,

and the exterior angle FDC is equal to the interior and opposite angle EAB; (1. 29.)

therefore the base FC is equal to the base EB. (I. 4.)
and the triangle FDC is equal to the triangle EAB.
From the trapezium ABCÊ take the triangle FDC,
and from the same trapezium take the triangle EAB,
and the remainders are equal, (ax. 3.)

therefore the parallelogram ABCD is equal to the parallelogram EB CF. Therefore, parallelograms upon the same, &c. Q.E.D.

PROPOSITION XXXVI. THEOREM.

Parallelograms upon equal bases and between the same parallels, are equal to one another.

Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG.

Then the parallelogram ABCD shall be equal to the parallelogram EFGH.

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Then because BC is equal to FG, (hyp.) and FG to EH, (1. 34.) therefore BC is equal to EH; (ax. 1.)

and these lines are parallels, and joined towards the same parts by the straight lines BE, CH;

but straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves equal and parallel; (I. 33.)

therefore BE, CH are both equal and parallel;
wherefore EBCH is a parallelogram. (def. A.)

And because the parallelograms ABCD, EBCH, are upon the same base BC, and between the same parallels BC, AĦ; therefore the parallelogram ABCD is equal to the parallelogram EBCH. (1.35.)

For the same reason, the parallelogram EFGH is equal to the parallelogram EBCH;

therefore the parallelogram ABCD is equal to the parallelogram EFGH. (ax. 1.)

Therefore, parallelograms upon equal, &c. Q.E.D.

PROPOSITION XXXVII. THEOREM.

Triangles upon the same base and between the same parallels, are equal to one another.

Let the triangles ABC, DBC be upon the same base BC,
and between the same parallels AD, BC.

Then the triangle ABC shall be equal to the triangle DBC.

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Produce AD both ways to the points E, F;
through B draw BE parallel to CA, (I. 31.)
and through C draw CF parallel to BD.

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Then each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DBCF, (1. 35.) because they are upon same base BC, and between the same parallels BC, EF. And because the diameter AB bisects the parallelogram EBCA, therefore the triangle ABC is half of the parallelogram EBCA; (1. 34.) also because the diameter DC bisects the parallelogram DBCF, therefore the triangle DBC is half of the parallelogram DBCF, but the halves of equal things are equal; (ax. 7.) therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles &c.

Q. E.D.

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