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NOTES.

A theorem is a proposition, wherein something is proposed to be demonstrated.

A problem is a proposition, wherein something is to be done or effected.

A lemma is some demonstration, previous and necessary, to render what follows the more easy.

A corollary is a consequent truth, deduced from a foregoing demonstration.

A scholium, is a remark or observation made upon something going before.

THE SIGNIFICATION OF SIGNS.

The sign=, denotes the quantities between which it stands to be equal.

The sign +, denotes the quantity it precedes to be added.

The sign, denotes the quantity which it precedes to be subtracted.

The sign X, denotes the quantities between them to be multiplied into each other.

To denote that four quantities. A, B, C, D, are proportional, they are usually written thus, A:B::C:D; and read thus, as A is to B, so is C to D; but when three quantities A, B, C, are proportional

, the middle quantity is repeated, and they are written A:B :: B: C.

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GEOMETRICAL THEOREMS.

THEOREM I.

Plate I.

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.

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

2. But if EB fall slantwise on CD, then are the angles DBE

+ EBC = DBE + EBA (= DBA ABC, or two right angles. Q. É. D.

Corollary 1. Whence if any numbers 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 lines.

+

2. And all the angles which can be made about a point, will be equal to four right angles.

THEOREM II.

Plate 1.

If one right line cross another, (as AC does BD) the opposite angles made by those lines, will be equal to each other ; that is AEB to CED and BEC to AED. fig. 21. By theorem 1. BEC + CED = 2 right angles.

and CED + DEA= 2 right angles.

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

After the same manner CED + AED= 2 right angles; and AED + AEB = two right angles ; wherefore taking AED from both, there remains CED = AEB. Q. E. D.

THEOREM III.

If a right line cross two parallels, as GH does AB and CD (fig. 22.) then,

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.

Plate I.

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 considered 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 FEB= EFC.

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 XEF + CFE are equal to two right angles. Q. E. D.

THEOREM IV.

In any triangle ABC, one of its legs, as BC, being produced towards D, it will make the external angle ACD equal to the two internal opposite angles taken together. Viz. to B and A, fig. 23.

Plate I.

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 ACE=A (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 CB 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.

Cor. 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.

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