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But it was proved that the 2 BFL is equal to the <FLC.

Therefore the EF A is also equal to the < FLC.

In a similar way it might be shown that the <HLD is equal to the ZL FB, and the Z HLC to the ZLFA.

PROPOSITION XXX. Straight lines which are parallel to the same straight line are parallel to each other.

To prove this proposition we must know,-
1. That magnitudes which are equal to the same

are equal to one another. (Ax. I.)
2. That if two straight lines are parallel, and are

cut by a third straight line, the alternate 8 80

formed are equal. (Prop. XXIX.). 3. That if two straight lines are parallel, and are

cut by a third, the exterior _ is equal to the interior and opposite < on the same side of the

intersecting line. (Prop. XXIX.) 4. That if two straight lines are crossed by a

third, and the exterior _ so formed is equal to the interior and opposite on the same side of the intersecting line, those straight lines are

parallel. (Prop. XXVIII.) Case 1. Suppose the two straight lines A B and CD are each parallel to EF. We have to prove that A B and CD are parallel to each other.

Proof. Draw the straight C line GM, intersecting the lines A B, CD, and EF, in the points H, K, and L. Because the lines A B and EF are parallel and the line GM cuts them, the

G

H

A

B

D

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E

M

G

H

exterior _GHA is equal to the interior and opposite ZHLE.

Again, because CD and EF are parallel, the exterior < HKC is equal to the interior and opposite < KLE.

Therefore, since the _sGH A and HKC are both equal to the < KLE, they are equal to each other.

Now the <8 GH A and HKC are the exterior and interior opposite <s, formed by the intersection of the two lines A B and C D by the line G M.

But if two straight lines are crossed by a third, and the exterior 2 so formed is equal to the interior and opposite < on the same side of the intersecting line, those straight lines are parallel. (Prop. XXVIII.) Therefore the lines A B and C D are parallel.

Case 2. Suppose the lines A B and CD are both parallel to the line EF B (which lies between them).

Draw the line GM intersecting the three lines D in the points H, K, and L.

Because the lines AB

and EF are parallel, and the line GM intersects them, the exterior _ GHA is equal to the interior and opposite <HLE.

Again, because EF nd C D are parallel, and the line GM intersects them, the exterior < HLE is equal to the interior and opposite _ LKC.

Therefore, since the <s GH A and LKC are both equal to the <HLE, they are equal to each other.

We have, therefore, two straight lines, A B and CD, cut by a third, so that the exterior _ thus formed (namely G HA) is equal to the interior and opposite < on the same side of the intersecting line (namely, the HKC).

A

E

F

к

с

M

It follows, therefore, according to the 28th proposition that these two right lines A B and C D are parallel to each other.

PROPOSITION XXXI,

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

For the construction requisite in this proposition

we must be able,1. To join two given points by a straight line.

(Post. I.) 2. From a given point in a given straight line to

draw a straight line making with the given line

an equal to a given Z. (Prop. XXIII.) To prove that the construction effects what is re

quired we must know,That if two straight lines are met by a third, and

the alternate s so formed are equal to one another, those two straight lines are parallel.

(Prop. XXVII.) Let AB be the given straight line, and D the given point. We have to draw through the point D a line parallel to the given line A B.

In the line AB take any point, as C.

Join D and C by the straight line DC.

From the point D draw the straight line D E, making with the line DC an _, CDE, equal to the _DCB. (Prop. XXIII.)

Produce the line E D through the point D to any point F. EF is the line required.

For since the two straight lines E F and A B are

E

F

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А

B

с

met by a third straight line, and the alternate Z8 80 formed, namely, CDE and DCB, are equal, the lines E Fand A B are parallel. (Prop. XXVII.)

PROPOSITION XXXII.

a

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

For the construction employed in this proposition

we must be able to draw through a given point a right line parallel to a given right line.

(Prop. XXXI.) For proving the proposition by the aid of the con

struction we must know,1. That if the same, or equal quantities be added

to equals, the sums will be equal. (Ax. II.) 2. That if one straight line meet another, the

adjacent Z8 so formed are together equal to

two right 8. (Prop. XIII.) 3. That if two straight lines are parallel and a

third straight line meets them, the alternate Zs so formed are equal, and the exterior Z is equal to the interior and opposite on the same side

of the intersecting line. (Prop. XXIX.) Let A B C be a A, of which one side, A C, has been

produced.

It has to be shown that the exterior _ BCD is equal to the sum

of the two interior and opposite DZS, BAC and ABC, and that

the three interior Zs of the A A B C are together equal to two right Zs.

B

E

A

с

Through the point C draw the straight line CE parallel to the side A B.

The exterior < BCD is thus divided into two parts; and the proof of the proposition consists in showing that one of these parts is equal to one of the interior and opposite Zs; and the other part to the other.

Since the lines A B and C E are parallel and are met by a third line, B C, it follows (Prop. XXIX.), that the alternate <s so formed, namely A B C and BCE, are equal.

Again, since A B and C E are parallel, and are intersected by a third line, A D, it follows that the exterior _ ECD is equal to the interior and opposite < on the same side of the intersecting line, namely the < B A C.

But if equals be added to equals, the sums are equal. (Ax. II.)

Therefore, if the equal <s A B C and BCE be added respectively to the Zs BAC and ECD, the sums will be equal: that is, the sum of the Zs BOE and E C D will be equal to the sum of the Zs ABC and B A C.

B C E and ECD together make up the exterior < BOD.

Therefore the exterior <BCD is equal to the sum of the two interior and opposite Z8, A B C and BAC.

Now, if to each of these equals be added the < B C A, the sums will be equal. (Ax. II.)

That is to say, the sum of the exterior _ BCD, and the < BCA, will be equal to the sum of the s ABC, B AC, and BCA

But since the straight line B C meets the straight line A D, the sum of the adjacent <s B C D and B C A is equal to two right Zs.

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