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sth. In the fourteenth, we say at different sides, for it they met at tlie same they could not lie in directum. For it is evident that two right lines iniglit meet another at the same point, and make angles (with it together – to two right angles (for one of them might be obtuse as the uther is acute) and not be in directum, if they both be at the same side of the line which they meet.
9th. In the fifteenth, by supplement is meant what either of the vertically opposite ongles wants of two right angles.
10th. In the twentieth, the difference between two sides is Jess than the third ; for if it was = then one side would be = to the other two;, and if greater one side would be greater than the other two.
11th. In the twenty-fourth, we say with the side which is not the greater, not with the lesser, because they might be =.
We take this side to avoid the different cases, taking the other might involve, as if the otber was taken, then the extremity of the new drawn line might fall above, below, or upon the base; but by taking the side not the greater it must fall below'it; as is proved in the notes to Dr. Elrington's Euclid.
lan. In Cor. 7, of thirty-second, we must not produce the sides consecutively, i. e. no two of them are to be produced from the common extremity.
13th. Conterminous means having a common point of termination.
14tb. Another way for shewing the equality of triangles is when they have two sides respectively = and the coue tained angles supplemental.
15th. Wheo two triangles have two sides mutually = but the contained angles either greaser or less than two right angles, that triangle will have the greater area whose angle (contained by those sides) approaches nearer to a right angle. This is evident from prop. 15 of deducibles.
16th. The distinction between coincident and = tria angles is : coincident triangles are those whose sides and angles are mutually =; and = triangles those whose areas only are 3.
17th. 'If we bisect the diagonal of a parallelogram and draw a line through the point of bisection, this line will bisect the parallelogram.
18th. 'i'be difference between a problem and a theorein is, that a problem consists of the data and qæsita and requires solution; a Theorem consists of the hypothesis and the thing asserted, and requires demonstration.
19th. In the third, the first leg produced is the leg of the triangle wbich terminates at the first centre, the second is that which terminates at the given point.
If the point be so situated that the lines drawn from it to the extremity of the base form an equilateral triangle the thing is done ; for then either of them is = to the given line.
If the point is in the line, then with the part of the line intercepted between the point and either extremity of the given line form an equilat. tri. and proceed as before.
If the point occurs in the extremity of the given line, it is not necessary to construct an equilat, tri
. but from the given point as centre and given line as radius, describe a circle ; any line drawn from the given point to meet the periphery of this will be = to the given line.
TABLE OF SIGNS.
Sign of Equality.
THE MOST IMPORTANT
Problems and Theorems, ,
FROM THE FIRST BOOK.
PROP. I, PROB.
To draw from a point (A) given in the side of a triangle
(DEF) a right line bisecting said triangle.
ANALYSIS. Suppose it to be done, and that the right line AB bisects the given triangle. Bisect the side ĎF in C (10. J. of Elr.) join AE, CE and CB. Then the triangles FEC and CED are = (38. 1. of Elr.) .. FEC is half DEF and is consequently = to FBA, if FBC be taken away; the remaining triangles BAC and CBE are =, and they being = and standing on the same base must be between the same parallels (39. 1. Elr.) Therefore, it is evident, that if you bisect DF in C join AE and draw CB parallel to it, that the straight line drawn from the given point-A to B, bisects the triangle.
For, join CE. then the triangles CEB and ABC are = 27. 1. Elr.) , if you add the common part FCB to both, the whole BAF is = to the whole EFC, .. &c.
PROP. 2, PROB.
Ta draw through a given point (P) a right line making
equal angles with two lines (AB and AC) given in position.
ANALYSIS. Suppose it done, and that GF is the required line. Draw through the external angle EAC, AK parallel to GF (31. ), of Elr.) Then the alternate angles KAD, ADF are = (29. 1. Elr.) and also the external angle EAK is = to tlie internal AFD,.. the angles EAK, KAC are =, and the line KA bisects the whole angle EAC.
Therefore it is evident if you bisect the angle EAC by AK and draw through D, GF parallel to AK, that the angles AFD and ADF, are =
For ADF and DAK are = and EAK and AFD are also = (29. I. Er.) ::. ADF and AFD are =.
PROP. 3. PROB,
From a point (C) given in the side of a parallelogram
ADÈB to draw a line bisecting it.
F draw the diagonal B!). Then the triangle DBE is half • the parallelogram (by 31. 1. of Elr.) also the figure FCBE
is half of it .. if you take away the common figure FGBE the triangles DGF and BGC are = it is also evident that they are equiangular .. they are equilateral, .. the diagonal is bisected by the line CF. Then if you bisect bisect the diagonal and draw from the given point, a line through the point of bisection it bisects the parallelogram,
PROP. 4, PROB,
To draw to the same point, in a right line AB given in
position tro equal right lines from two given points (D, E) without it.
lines required to be drawn; connect the given points, it is evident that the angles FDE,FED are =, bisect the angle DFE by FC, then it is evident that the angles FCE, FCD are = and are ., right angles, and also that DC is = to CE (by 4. 1. Elr.)
Then it is evident, that if the points are connected, the connecting line bisected and a perpendicular be drawn from the point of bisection to nieet the given line, the point in which it meets it is the point to which the lines can be drawn.
Note. The points must be so situated that the line connecting them cannot be at right angles to given line,
PROP. 5, THEOR.
Equal triangles (ABC, DEF) between the same pa
1 allels stand on equal bases. Fig. 6.
For, if they do not, cut from the greater BC a part GC to EF, connect GA.
Then the triangles GAC, EDF are = (38. 1. Elr.). GAC is = to ABC, a part to the whole which is absurd.
PROP. 6. THEOR.
If one side (AB) of any triangle be bisected, and a right
line (DF) be drann ihrough the point of bisectiun (D) parallel to the base it bisects the third side.
ANALYSIS. Suppose it to be the case ; connect FB, then since AF is = to FC the triangles ABF and FBC are =;..if FBC is proved = to balf of ABC the theorem is proved.
Join DC, since AD is = to DB the triangles ADC and BDC are = .. BDC is half ABC ; but BDC and CFB are = since they are on the same base and between the same par. .:: BFC is half ABC and .. (by the foregoing theur.) AC is bisected by DF.
PROP. 7, THEOR.
If a right line (DF) bisects any two sides of a triangle
it is parallel to the third sile. And the triangle (ADF) cut off by it, is the fourth part of the rehole triangle.