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HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES.
It is commonly found that exercises in Pure Geometry present to a beginner far more difficulty than examples in any other branch of Elementary Mathematics. This seems to be due to the following
(i) The variety of such exercises is practically unlimited ; and it is impossible to lay down for their treatment any definite methods, such for example as the rules of Elementary Arithmetic and Algebra.
(ii) The arrangement of Euclid's Propositions, though perhaps the most convincing of all forms of argument, affords in most 'cases little clue as to the way in which the proof or construction was discovered.
Euclid's propositions are arranged synthetically: that is to say, starting from the hypothesis or data, they first give a construction in accordance with postulates, and problems already solved ; then by successive steps based on known theorems, they prove what was required in the enunciation.
Thus Geometrical Synthesis is a building up of known results, in order to obtain a new result.
But as this is not the way in which constructions or proofs are usually discovered, we draw the student's attention to the following hints.
Begin by assuming the result it is desired to establish ; then by working backwards, trace the consequences of the assumption, and try to ascertain its dependence on some simpler theorem which is already known to be true, or on some condition which suggests the necessary construction. If this attempt is successful, the steps of the argument may in general be re-arranged in reverse order, and the construction and proof presented in a synthetic form.
This unravelling of a proposition in order to trace it back to some earlier principle on which it depends, is called geometrical analysis : it is the natural way of attacking many theorems, and it is especially useful in solving problems.
Although the above directions do not amount to a method, they often furnish a mode of searching for a suggestion. Geometrical Analysis however can only be used with success when a thorough grasp of the chief propositions of Euclid has been gained.
The practical application of the foregoing hints is illustrated by the following examples.
1. Construct an isosceles triangle having given the base, and the sum of one of the equal sides and the perpendicular drawn from the vertex to the base.
Let AB be the given base, and K the sum of one side and the perpendicular drawn from the vertex to the base. ANALYSIS. Suppose ABC to be the required triangle.
From C draw CX perpendicular to AB :
then AB is bisected at X.
it follows that CH=CA;
and if AH is joined, we notice that the angle CAH=the angle CHA. Now the straight lines XH and AH can be drawn before the position of C is known ;
Hence we have the following construction, which we arrange synthetically. SYNTHESIS.
Bisect AB at X: from X draw XH perpendicular to AB, making XH equal to K.
1. 4. Again, since the angle HAC=the angle AHC, Constr. .. HC=AC.
1. 6. To each add CX; then the sum of AC, CX=the sum of HC, CX
=HX. That is, the sum of AC, CX=K.
2. To divide a given straight line so that the square on one part may be double of the square on the other.
Let AB be the given straight line.
ANALYSIS. Suppose AB to be divided as required at X: that is, suppose the square on AX to be double of the square on XB.
Now we remember that in an isosceles right-angled triangle, the square on the hypotenuse is double of the square on either of the equal sides.
This suggests to us to draw BC perpendicular to AB, to make BC equal to BX, and to join XC. Then the square on XC is double of the square on XB; I. 47.
1. 5. Thus the exterior angle CXB is double of the angle XAC. But the angle CXB is half of a right angle :
1. 32. :: the angle XAC is one-fourth of a right angle.
This supplies the clue to the following construction :-
and from A draw AC, making BAC one-fourth of a right angle. From C, the intersection of AC and BD, draw CX, making the angle ACX equal to the angle BAC.
1. 23. Then AB shall be divided as required at X. For since the angle XCA= the angle XAC, 1. XA=XC.
1. 6. And because the angle BXC=the sum of the angles BAC, ACX, 1. 32.
.. the angle BXC is half a right angle.
And the angle at B is a right angle;
BX=BC. Hence the square on XC is double of the square on XB : 1. 47. that is, the square on AX is double of the square on XB. Q.E.F.
ON THE IDENTICAL EQUALITY OF TRIANGLES.
See Propositions 4, 8, 26.
1. If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles.
2. If the bisector of the vertical angle of a triangle is also perpendicular to the base, the triangle is isosceles.
3. If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles.
[Produce the bisector, and complete the construction after the manner of 1. 16.]
4. If in a triangle a pair of straight lines drawn from the extremities of the base, making equal angles with the remaining sides, are equal, the triangle is isosceles.
5. If in a triangle the perpendiculars drawn from the extremities of the base to the opposite sides are equal, the triangle is isosceles.
6. Two triangles ABC, ABD on the same base AB, and on opposite sides of it, are such that AC is equal to AD, and BC is equal to BD: shew that the line joining the points C and D is perpendicular to AB.
7. If from the extremities of the base of an isosceles triangle perpendiculars are drawn to the opposite sides, shew that the straight line joining the vertex to the intersection of these perpendiculars bisects the vertical angle.
8. ABC is a triangle in which the vertical angle BAC is bisected by the straight line AX : from B draw BD perpendicular to AX, and produce it to meet AC, or AC produced, in E ; then shew that BD is equal to DE.
9. In a quadrilateral ABCD, AB is equal to AD, and BC is equal to DC: shew that the diagonal AC bisects each of the angles which it joins.
10. In a quadrilateral ABCD the opposite sides AD, BC are eqnal, and also the diagonals AC, BD are equal: if AC and BD intersect at K, shew that each of the triangles AKB, DKC is isosceles.
11. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle,
12. Two right-angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects.
Let ABC, DEF be two As right-angled at B and E, having AC equal to DF, and AB equal to DĚ.
Then shall the A ABC be equal to the A DEF in all respects.
For apply the A ABC to the A DEF, so that AB may coincide with the equal line DE, and C may fall on the side of DE remote from F. Let C' be the point on which C falls. Then DEC' represents the A ABC in its new position. Now each of the__8 DEF, DEC' is a rt. 2; :: EF and EC' are in one st. line.
1. 14. Then in the A C'DF, because DF=DC' (i.e. AC), Нур. :: the LDFC'=the L DC'F.
I. 5. Hence in the two A8 DEF, DEC',
the L DEF=the L DEC', being rt. 28; Because and the LDFE=the L DC'E;
Proved. also the side DE is common to both; the As DEF, DEC' are equal in all respects; that is, the AS DEF, ABC are equal in all respects. Q.E.D.
Alternative Proof. Since the L ABC is a rt. angle ; .. the sq. on AC=the sqq. on AB, BC.
I. 47. Similarly, the sq. on DF=the sqq. on DE, EF ;
I. 47. But the sq. on AC=the sq. on DF, since AC=DF;
: the sqq. on AB, BC=the sqq. on DE, EF. And of these, the sq. on AB=the sq. on DE, since AB=DE ; : the sq. on BC=the sq. on EF ;
Ax. 3. :: BC=EF. Hence the three sides of the ABC are respectively equal to the three sides of the A DEF;
:: the A ABC=the A DEF in all respects. 1. 8.