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PROP. XXXI.-PROBLEM.

To draw a straight line through a given point, parallel to a given straight line.

(References—Prop. I. 23, 27.) Let A be the given point, and BC the given right line. It is required to draw a straight line through the point A, parallel

to BC.

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In BC take any point D, and join AD; at the point A, in the straight line AD, make the angle DAE equal to

the angle ADC; (L. 23) and produce the straight line EA to F.

Then EF shall be parallel to BC.

DEMONSTRATION

Because the straight line AD, which meets the two straight lines

BC, EF, makes the alternate angles EAD, ADC, equal to one another,

therefore EF is parallel to BC. (I. 27.) Wherefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.

Q. E. F.

PROP. XXXII.-THEOREM.

If the side of any triangle be produced;

then 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.

(References-Prop. 1. 13, 29, 31; ax. 2.) Let ABC be a triangle, and let one of its sides BC be produced to D. Then the exterior angle ACD is equal to the two interior

and opposite angles CAB, ABC;

and the three interior angles of the triangle - viz., ABC,

BCA, CAB, are together equal to two right angles.

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Through the point C draw CE parallel to the straight line AB. (1. 31.)

DEMONSTRATION

Because AB is parallel to CE, and AC meets them,

therefore the alternate angles BAC, ACE, are equal. (1. 29.) Again, because AB is parallel to CE, and BD falls upon them, therefore the exterior angle ECD is equal to the interior and opposite

anyle ABC; (I. 29) but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior anglo ACD is equal to the

two interior and opposite angles CAB, ABC. (ax. 2.)

To each of these equals add the angle ACB; therefore the angles ACD, ACB, are equal to the three angles CBA,

BAC, ACB; but the angles ACD, ACB, are equal to two right angles; (1. 13) therefore also the angles CBA, BAC, ACB, are equal to

two right angles. (ax. 1.) Wherefore, if a side of a triangle, &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.

D

E

DEMONSTRATION

For, any rectilineal figure ABCDE can, by drawing straight lines from

a point F within the figure to each angle, be divided into as many triangles as the figure has sides.

By the proposition, the angles of each triangle are equal to two right

angles; therefore all the angles of the triangles are equal to twice as many

right angles as there are triangles, that is, as there are sides of the

figure ; but the same angles are equal to the angles of the figure, together with

the angles at the point F; and the angles at F, which is the common vertex of all the triangles,

are equal to four right angles, (I. 15, cor. 2) 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 are together equal to four right angles.

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For, because each interior angle ABC, together with its adjacent ex

terior angle ABD, are equal to two right angles, (I. 13) therefore all the interior angles, together with all the exterior angles,

are equal to twice as many right angles as there are angles or

sides. but all the interior angles, together with four right angles, are equal to

twice as many right angles as the figure has sides; (cor. 1) 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 away the interior angles which are common;

therefore all the exterior angles are equal to four right angles. (ax. 3.) PROP. 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

(References-Prop. I. 4, 27, 29.) Let AB, CD, be two 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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Because AB is parallel to CD, and BC meets them,

therefore the alternate angles ABC, BCD, are equal; (1. 29) and because AB is equal to CD, (hyp.) and BC common to the two

triangles ABC, DCB, the two sides AB, BC, are equal to the two DC, CB; and the angle ABC has been shown 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 shown to be equal to BD. Therefore, straight lines, &c.

Q. E. D.

PROP. XXXIV.—THEOREM. The opposite sides and angles of parallelograms are equal to each other, and the diameter bisects them, that is, divides them into two equal parts. N.B. A parallelogram is a four-sided figure, of which the opposite sides are parallel; and the diameter is the straight line joining two of its opposite angles.

(References-Prop. I. 4, 26, 29; ax. 2.)

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 shall bisect it.

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DEMONSTRATION Because AB is parallel to CD, and BC meets them, therefore the alternate angles ABC, BCD, are equal to one another ;

(1. 29) and because AC is parallel to BD, and BC meets them,

therefore the alternate angles ACB, CBD, are equal to one another. Wherefore 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, is common

to the two triangles ; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other

(1. 26)—viz., 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 shown to be equal to the angle BDC; therefore the opposite sides and angles of a parallelogram

are equal to one another. Also the diameter shall bisect it. For because AB is equal to CD, and BC common, the two AB, BC, are equal to the two DC, CB, each to each ; and the angle ABC has been proved equal to the angle BCD;

therefore the triangle ABC is equal to the triangle BCD; (I. 4) and the diameter BC divides the parallelogram ACDB into two equal parts.

Q. E. D.

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