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Wherefore a triangle KFG has been made, having its three sides equal to the three lines A, B, C.
Q. E. F.
Proof (with contracted Syllogisms, in the words of
Because the point F is the centre of circle
... FD is equal to FK.
But FD is equal to A.
(1).. F K is equal to A.
Again, the point G is the centre of circle
.. G H is equal to G K. (Definition.)
But G H is equal to C. (Construction.)
(2).. G K is equal to C.
(3) And FG was made equal to B. (Construction.)
Result... The three straight lines KF, FG, GK are equal to the three sides A, B, C.
Wherefore a triangle KFC has been made, having its three sides equal to the three given lines A B C. Q. E. F. EXERCISES.-I. Write out the last problem without referring to the book, with the contracted proof, using figures instead of letters.
II. On the given straight line M
describe a triangle having its other two sides equal to the given lines P and Q.
(Use two circles, with the two given lines as radii, as in constructing an equilateral triangle.)
DEFINITION OF AN ANGLE.
A Rectilineal angle is the inclination (or sloping) of two straight
lines towards one another,
which meet (at a point),
but are not in the sarne
Thus, the two lines A
A B, A C slope towards one another, and meet at the point A, forming the angle BA C.
In naming an angle, we sometimes use the letter at the point where the lines meet, and say the angle at A.
But more generally we use three letters, placing that letter in the middle which stands at the point where the lines meet, and say the angle BA C, or the angle CA B.
The lines A B, A C are said to contain the angle BA C, and they may be produced ever so far in the direction of B and C without increasing the size of the angle.
To compare one triangle with another, we must apply one to the other.
Thus, to compare the angle B A C with the angle EDF; apply the angle B A C to the angle EDF;
so that the point A falls upon the point D, and the line AB on the line D E.
Then, if the line A C coincide with the line DF, the angles are equal.
If the line AC fall within the angle EDF, the angle B A C is less than ED F.
But if the line A C fall without the angle EDF, then the angle B A C is greater than ED F.
Try this with cardboard angles of various sizes. When several angles meet at a point, as in this
figure, be careful to distinguish one from the other by their proper letters.
Thus, at the point O, there are nine angles, viz. :—
1. The angle E OB. 3. The angle DOC. 5. The angle COB. 7. The angle DOE. 9. The angle AO C.
2. The angle COE.
The angle CO B would, in this case, be called the whole angle, consisting of the parts COE, E O B.
The angle DOB again might be spoken of as the whole angle, consisting of the parts DOC, COE, EOB.
EXERCISES.-I. State which of the nine angles named above are right angles, which acute, and which obtuse angles.
II. Put letters to the lines in this figure, and name all the angles meeting at the point O.
PROBLEM (Euclid I. 9).
Repeat. The definition of a rectilineal angle and of an equilateral triangle, and axiom 2a.
(Euc. I. 8.)-If two triangles have two sides of the one equal to two sides of the other, each to each, and have also their bases equal; then the angle contained by the two sides of one triangle shall be equal to the angle contained by the two sides of the other.1 (Assumed here as an axiom proved on page 76.)
To bisect a given rectilineal angle, that is to divide it into two equal parts.
1 The meaning of this must be made quite clear to the pupils by viva voce explanation.
If the angle B A C is bisected by AF; the angle DAF will be equal to the angle E A F.
.. We have to prove that angle DAF equals angle EAF.
Proof (with syllogisms in full).
ADF and AE F are two triangles (distinguish here between angles and triangles).
DA, A F are two sides of triangle A D F, and
EA, AF are two sides of triangle A E F, and
AD is equal to AE (by construction a);
add A F to each of these.
(c) Then, by Axiom 2α, AD, A F are equal to AE, AF.
All the sides of an equilateral triangle are equal. DF and EF are sides of an equilateral triangle (b). (d) ... DF and E F are equal.