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IF a right line falls on another, as AB, or EB, does on CD, (fig. 20.) it either makes with it two right angles, or two angles equal to two right angles.

B

1. If AB be perpendicular to CD, then (by def. 11, the angles CBA, and ABD, will be each & right angle,

2. But if EB fall slantwise on CD, then are the angles DBE+EBC=DBE+EBA (=DBA) +ABO, or to two right angles, . Q. E. D.

Corollary 1. Whence if any number of right lines were drawn from one point, on the same side of a right line; all the angles made by these lines will be equal to two right angles,

2. And all the angles which can be made about a

THEOREM II.

fig. 22.

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Is. If one right line cross another (as

IC does BD) the opposite angles
Rade by those lines, will be equal to
lach other; that is, AEB to CED
und BEC to AED. fig. 21.

A

By theorem 1. BEC+CED=2 right angles.

and CED+DEA=2 right angles.

Therefore (by axiom 1.) BEC+CED=CED+DEA: take CED from both and there remains BEC=DEA, (by axiom 5.) Q. E. D.

B

After the same manner CED+AED=2 right angles; and AED+AEB=2 right angles; wherefore, f. 1. taking AED from both, there remains CED=AER.

Q. E. D.

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G

I If a right line cross two paral- A ne lels, as GH does AB and CD

C (fig. 22. ) then, be

H

1. Their external angles are equal to each other, that is, GEB=CFH.

2. The alternate angles will be equal, that is, AEF =EFD and BEF=CFE.

3. The external angle will be equal to the internal and opposite one on the same side, that is, GEB= EFD and AEG=CFE.

4. And the sum of the internal angles on the same side, are equal to two right angles ; that is, BEF+ DFE are equal to two right angles, and AEF+CFE are equal to two right angles,

1. Since AB is parallel to CD, they may be consi, dered as one broad line, crossed by another line, as GH; (then by the last theo.) GEB=CFH, and AEG =HFD.

2. Also GEB=AEF, and CFH=EFD; but GEB =CFH (by part 1. of this theo.) therefore AEF= EFD. The same way we prove FEC=EFB.

3. AEF=EFD; (by the last part of this theo.) but AEF=GEB (by theo. 2.) Therefore GEB=EFD. The same way we prove AEG=CFE.

4. For since GEB=EFD, to both add FEB, then (by axiom 4.) GEB+FEB=EFD+FEB, but GEB+ FEB, are equal to two right angles (by theo 1.) Therefore EFD+FEB are equal to two right angles ; after the same manner we prove that AEF+CFE ere equal

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Thro’ C, let CE be drawn parallel to AB; then since BD cuts the two parallel lines BA, CE; the angle ECD--B (by part 3. of the last theo.) and again, since AC cuts the same parallels, the angle ACEEA (by part 2. of the last.) Therefore ECD+ACE=ACD =B+A. Q. E. D.

THEOREM V.

In any triangle ABC, all the three angles taken together are equal to two right angles, viz, A+B+ACB =2 right angles. fig. 23.

Produce BC to any distance, as D, then (by the last.) ACD=B+A; to both add ACB; then ACD+ ACB=A+B+ACB; but ACD+ACB=2 right angles (by theo. 1.) therefore the three angles A+B+ACB =2 right angles. Q. E. D.

Cer. 1. Hence if one angle of a triangle be known, the sum of the other two is also known: for since the three angles of every triangle contain two right ones, or 180 degrees, therefore 180—the given angle will be equal to the sum of the other two; or 180—the sum of two given angles, gives the other one.

Cor. 2. In every right-angled triangle, the two acute angles are:90 degrees, or to one right-angle : therefore 90-one acute angle, gives the other.

THEOREM VI.

f. 21.

D

If in any two triangles, ABC, DEF, there be two sides AB, AC in the one, severally equal to DE, DF in the other, and the angle A contained betroeen the two sides in the one, equal to D in the other; then the remaining angles of the one, will be severally equal to those of the other, viz. B-E and C=F: and the base of the one BC, will be equal to EF, that of the other. fig. 24.

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If the triangle ABC be supposed to be laid on the triangle DEF, so as to make the points A and B coincide with D and E, which they will do, because AB =DE (by the hypothesis); and since the angle A=D, the line AC will fall along DF, and inasmuch as they are supposed equal, C will fall in F; seeing therefore the three points of one coincide with those of the other triangle, they are manifestly equal to each other; therefore the angle E=E and C=F, and BC

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