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Note 1. In this Proposition the three sides of one triangle are given equal respectively to the three sides of the other; and from this it is shewn that the two triangles may be made to coincide with one another.
Hence we are led to the following important Corollary.
COROLLARY. If in two triangles the three sides of the one are equal to the three sides of the other, each to each, then the triangles are equal in all respects.
[An alternative proof, which is independent of Prop. 7, will be found on page 26.]
Note 2. Proposition 8 furnishes an instance of a true theorem of which the converse is not necessarily true.
It is proved above that if the sides of one triangle are severally equal to the sides of another, then the angles of the first triangle are severally equal to the angles of the second.
The converse of this enunciation would be as follows: If the angles of one triangle are severally equal to the angles of another, then the sides of the first triangle are equal to the sides of the second.
But this, as the diagram in the margin shews, is by no means necessarily true.
EXERCISES ON PROPOSITION 8.
1. Shew (by drawing a diagonal) that the opposite angles of a rhombus are equal.
2. If ABCD is a quadrilateral, in which AB=CD and AD=CB, prove that the angle ÅDC=the angle ABC.
3. If ABC and DBC are two isosceles triangles drawn on the same base BC, prove (by means of 1. 8) that the angle ABD=the angle ACD, taking (i) the case where the triangles are on the same side of BC, (ii) the case where they are on opposite sides of BC.
4. If ABC, DBC are two isosceles triangles drawn on opposite sides of the same base BC, and if AD be joined, prove that each of the angles BAC, BDC will be divided into two equal parts.
5. If in the figure of Ex. 4 the line AD meets BC in E, prove that BE=EC.
Let ABC and DEF be two triangles, which have the sides BA, AC equal respectively to the sides ED, DF, and the base BC equal to the base EF.
Then shall the angle BAC be equal to the angle EDF. For apply the triangle ABC to the triangle DEF, so that B may fall on E, and BC along EF, and so that the point A may be on the side of EF remote from D;
then C must fall on F, since BC is equal to EF. Let GEF be the new position of the triangle ABC.
CASE I. When DG intersects EF.
Then because ED=EG,
the angle FDG=the angle FGD.
that is, the angle EDF = the angle BAC.
Two cases remain which may be dealt with in a similar manner : namely,
QUESTIONS AND EXERCISES FOR REVISION.
1. Define adjacent angles, a right angle, vertically opposite angles. 2. Explain the words enunciation, hypothesis, conclusion. 3. Distinguish between the meanings of the following statements:
(i) then AB is equal to PQ ;
(ii) then AB shall be equal to PQ. 4. When are two theorems said to be converse to one another. Give an example.
5. Shew by an example that the converse of a true theorem is not itself necessarily true.
6. What is a corollary? Quote the corollary to Proposition 5 ; and shew how its truth follows from that proposition.
7. Name the six parts of a triangle. When are triangles said to be equal in all respects ?
8. What do you understand by the expression geometrical magnitudes ? Give examples ?
9. What is meant by superposition ? Explain the test by which Euclid determines if two geometrical magnitudes are equal to one another. Illustrate by an example.
10. Quote and explain the third postulate. What restrictions does Euclid impose on the use of compasses, and what problems are thereby made necessary ?
11. Define an axiom. Quote the axioms referred to (i) in Proposition 2 ; (ii) in Proposition 7.
12. Prove by the method of superposition that two squares are equal in area, if a side of one is equal to a side of the other.
13. Two quadrilaterals ABCD, EFGH have the sides AB, BC, CD, DA equal respectively to the sides EF, FG, GH, HE, and have also the angle BAD equal to the angle FEH. Shew that the figures may be made to coincide with one another.
14. AB, AC are the equal sides of an isosceles triangle ABC ; and L, M, N are the middle points of AB, BC, and CA respectively : (i) LM=MN.
(ii) BN=CL. (iii) the angle ALM=the angle ANM.
PROPOSITION 9. PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal parts.
Let BAC be the given angle.
It is required to bisect the angle BAC. Construction. In AB take any point D; and from AC cut off AE equal to AD.
Join DE; and on DE, on the side remote from A, describe an equilateral triangle DEF.
Constr. and AF is common to both; Because and the third side DF is equal to the third side
Def. 24. therefore the angle DAF is equal to the angle EAF. 1. 8. Therefore the given angle BAC is bisected by the straight line AF.
1. If in the above figure the equilateral triangle DFE were described on the same side of DE as A, what different cases would arise ? And under what circumstances would the construction fail ?
2. In the same figure, shew that AF also bisects the angle DFE. 3. Divide an angle into four equal parts.
PROPOSITION 10. PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts.
Let AB be the given straight line.
It is required to divide AB into two equal parts.
Then shall AB be bisected at the point D.
Def. 24. and CD is common to both ; Because
also the contained angle ACD is equal to the contained angle BCD;
Constr. therefore the triangle ACD is equal to the triangle BCD in all respects :
so that the base AD is equal to the base BD. Therefore the straight line AB is bisected at the point D.
1. Shew that the straight line which bisects the vertical angle of an isosceles triangle, also bisects the base.
2. On a given base describe an isosceles triangle such that the sum of its equal sides may be equal to a given straight line,