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COR. 1. If the two angles ACD, BCD, with the same vertex C', are together equal to two right angles, AC, and CB, are in one and the same straight line.

COR. 2. Hence, also, whatever be the number of angles in one plane, separate and distinct, on one side of AB, with a common vertex C, the sum of all these angles is equal to two right angles; and similarly on the other side of AB. So that all the angles in one plane exactly occupying the whole space round any given point are together equal to four right angles.

31. PROP. IX. If one straight line intersects* another straight line, the vertical, or opposite, angles shall be equal to one another.

Let the straight line AB intersect the straight line CD in the point E; then

LAEC = L BED,

and AED = ▲ BEC. For, since CE meets AB, and makes with it the angles AEC, BEC, these are together equal to two right angles (30); and since BE meets CD, and

D

makes with it the angles BEC, BED, these also are equal to two right angles; the angles AEC, BEC together are equal to the angles BEC, and BED together; and, taking the same angle BEC from these equals, the remainders must be equal, that is,

LAEC=L BED.

Similarly, < AEC, AED are equal to two right angles; and so also are 28 AEC, BEC;

.. LAED=L BEC.

COR. Hence, if the lines forming any angle be 'produced', or extended, through the vertex in the opposite direction, the new angle thus formed will be equal to the other.

32. PROP. X. If a side of a triangle be 'produced'†, ** One line intersects another when it not only meets that other, but crosses it and is continued on the other side.

+ That is, be extended, or continued indefinitely in the same straight line.

the exterior angle, thus formed by the adjacent side and the side produced, is greater than either of the interior opposite angles.

F

Let the side BC of the triangle ABC be 'produced' to any point D; the exterior angle ACD shall be greater than either of the interior opposite angles ABC, BAC.

Bisect the side AC in the point E (27), join

BE, produce it to F,

making EF equal to EB B

D

(by describing a circle with centre E and radius EB to cut BE produced in F), and join CF.

Then in the triangles AEB, CEF, AE, EB are equal to CE, EF, each to each, by construction, and 2 AEB = =▲ CEF (31), because they are opposite vertical angles, the triangles AEB, CEF are equal in all respects (24), and the angles are equal to which the equal sides are opposite, so that BAE= ECF; but ‹ ACD is greater than 4 ECF (8), .. LACD is greater than L BAE, or BAC, which is the same thing.

Similarly it may be shewn, by bisecting BC in G, joining AG, and proceeding as before, that ACD is greater than ABC.

L

33. PROP. XI. The greater side of every triangle is opposite to the greater angle.

Let ABC be a triangle, of which the side AC is greater than the side AB;

ABC shall be greater

than BCA.

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With centre A and radius AB describe an arc of a circle cutting AC in the point D; and join BD,

which will necessarily fall within the triangle ABC.

Then, since AB = AD, LABD = 4 ADB (26); but LADB, the exterior angle of the triangle DBC, is greater than the interior opposite BCD (32); .. LABD is greater than BCD, or 4 BCA, and .. à fortiori 4 ABC is greater than 4 BCA.

L

COR. Conversely, the greater angle of every triangle is subtended by the greater side.

34. PROP. XII. If a straight line, meeting two other straight lines in the same plane, form angles with each, on contrary sides of itself, equal to one another, these two straight lines shall be parallel.

[The angles here described are, for shortness, called alternate angles.]

Let the straight lines AB, CD, be met by the straight line EF, so that AEF:

LEFD; or BEF = < CFE ; then AB is parallel to CD.

For, if AB, CD are not parallel, they will meet, upon being produced either towards B, D, or towards A, C. Sup- a pose them to meet towards B, D, then a triangle will be

formed of which EF is one

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G

D

side, and AEF will be the exterior' angle of that triangle, mentioned in a former proposition (32), .. 4AEF is greater than the interior opposite angle EFD. But, by the supposition, < AEF = 2 EFD; and.. if AB, CD meet anywhere towards B, D, AEF is both greater than, and equal to, EFD; which is manifestly impossible. Hence AB, CD do not meet towards, B, D; and in the same manner it may be shewn that they do not meet towards A, C; .. AB, CD are parallel, according to the Definition of parallel lines (12).

COR. 1. Produce FE to any point G; if the exterior angle, as BEG, be equal to the interior and opposite angle on the same side of the intersecting line, as EFD; then also AB shall be parallel to CD.

For BEG. = LAEF (31), .: LAEF = EFD, and .. AB, CD are parallel, as already proved.

COR. 2. If the two interior angles on the same side of the intersecting line are together equal to two right angles, the two straight lines shall be parallel.

For, since BEF + BEG = two right angles (30), .. BEF+ ‹ BEG = ▲ BEF + ▲ EFD,

and ... BEG=EFD, .. AB is parallel to CD, by COR. 1. COR. 3. If two straight lines be parallel, and from

any point in one of them a straight line be drawn at right angles to that one, and produced to meet the other, it will also be at right angles to the other. This is obvious from Cor. 2.

COR. 4. The converse of both the original proposition and each of the three preceding Corollaries also holds true, viz. that if two parallel straight lines be intersected by a third, the alternate angles are equal to one another; and the exterior angle is equal to the interior and opposite upon the same side; and also the two interior angles upon the same side are equal to two right angles.

35. PROP. XIII. The straight line which joins two parallel straight lines, and is at right angles to each of them, is the shortest line which can be drawn from one to the other. Let AB, CD be two parallel straight lines, take any point E in AB, and draw EF at right angles to AB, meeting CD in F; EF shall be the shortest line which can be drawn from E to meet

E

F

B

CD. From E draw any other line EG to meet CD in some point G between F and D. Then BEG = LEGF (34), since they are 'alternate angles'; and L BEF is greater than BEG, .. EFG is greater than EGF. But the greater side is opposite to the greater angle, (33. Cor.) EG is greater than EF, and EG is any other line than EF joining the parallels, .. EF is the shortest of all such lines.

...

COR. All lines joining at right angles the same two parallels are equal to each other. The most common and popular notion of parallel lines is based upon this property.

36. PROP. XIV. To draw a straight line parallel to a given straight line through any proposed point without it. Let AB be the given straight line, and C the given point without it. It is required to draw, through the point C, a straight line parallel to AB.

H

Take any point D in
AB, and join CD; bi- A

D

sect CD in the point E (27); from E draw EF to meet AB in any other point F; produce FE to G, making EG=EF (by drawing a circle with centre E and radius EF); join the points G and C by the straight line GC, and produce GC both ways indefinitely to H and I; then HCI is a straight line through the point C parallel to AB.

=

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=

For ED EC, and EF EG, . in the triangles EDF, ECG, the two sides ED, EF are equal to the two sides EC, EG; also DEF = 4 CEG (31), .. the triangles are equal in all respects, and EDF: = 4 ECG, or, which is the same thing, CDB = 4 HCD, and they are 'alternate angles', .. HCI is parallel to AB (34).

Another Method. The same thing may be done as follows:

From C draw CD perpendicular to AB (29), and again from C draw CE at right E

angles to CD (28); produce

EC to any point F; then EF is parallel to AB and is drawn through C (34, Cor. 3).

A

C

D

F

B

37. PROP. XV. If a side of any triangle be produced, the exterior angle (formed by the adjacent side and the side produced) is equal to the two interior opposite angles of the triangle; and the three interior angles of every triangle are equal to two right angles.

Let the side AB of the triangle ABC be 'produced' to D; CBD

L

shall be equal to the angles BAC, ACB taken together; and

the three angles

ABC, ACB, BAC,

shall together be

equal to two right 4 angles.

B

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D

Through the point B draw BE parallel to AC (36); then because BC meets the two parallel straight lines AC, BE, LEBC= ACB, and EBD = BAC, (34, Cor. 4); .. CBD, which is made up of EBC, and 2 EBD, is equal to the two interior opposite angles BAC, ACB.

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