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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.
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
Is. If one right line cross another (as
IC does BD) the opposite angles
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.
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.
I If a right line cross two paral- A ne lels, as GH does AB and CD
C (fig. 22. ) then, be
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
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.
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.
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.
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