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Produce DA, DC to E and F. as before.)
II. Let M be the given point, and NO the given line: draw a line from M equal to N O.
Then draw the circles,
III. Let X be the given point, and YZ the given line: draw two lines from X equal to Y Z.
(Join XZ and X Y, and construct two equilateral triangles.)
IV. Let (1.) be the given point, and (2, 3) the given line:
draw two lines from (1.) equal to (2, 3).
Theorems of Euclid, assumed as true for the purposes of proof in Part I., but proved in Part II.
Euclid I. 4.-If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each-namely, those to which the equal sides are opposite. (For proof see page 70, Part II.)
Euc. I. 8.-If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides, equal to them of the other. (For proof see page 76, Part II.)
METHODS OF DEMONSTRATION.
The proof of a proposition may be either direct or indirect.
In a direct demonstration we prove directly the truth of an assertion.
All the problems worked out in the first part are proved by direct demonstration.
In proving the truth of a theorem by direct demonstration, we proceed sometimes by construction, sometimes by superposition.
We proceed by construction, when we draw certain lines, by the aid of which we prove the truth of the assertion.
We proceed by superposition, when we suppose one figure to be placed upon another, and having shown that each line of the one figure lies exactly upon the corresponding line of the other figure, infer from that the equality of the figure.
(The proposition in the article on "A Chain of Syllogisms," page 16, is proved by superposition.)
In an indirect demonstration, we do not prove that the assertion is true, but that it is not untrue.
Thus, in Euc. I. 6, we are asked to prove that if the angles at the base of a triangle are equal, the two sides must be equal; we do not prove that directly, but we prove that it is impossible they should be unequal.
We prove that an assertion cannot be untrue, by a method which is called the reductio ad
That is, by reducing to an
In using the reductio ad absurdum we suppose the statement to be untrue, and then show that some absurd result follows from our supposition.
Thus, in Euc. I. 6, as above, to prove the two sides equal, we suppose them to be unequal. And then we find that, supposing the sides to be unequal, we are able to prove that a small triangle is equal to a larger triangle. As this is manifestly absurd, the supposition on which it is founded must also be absurd, and therefore untrue.
A somewhat different kind of indirect demonstration is found in such propositions as Euc. I. 14, where we are asked to prove that certain two straight lines are in one and the same straight line.. We do not prove that. But we do prove that no other straight line but the one in question can possibly be in a straight line with the second line. And hence, by inference, that particular line must be.
We now proceed to prove several of Euclid's theorems by one or other of these methods.
THEOREM (Euclid I. 13).
Repeat. The difference between a problem and a theorem; the definition of a right angle; and Axioms 1 and 2a.
The angles which one straight line makes with another on one side of it are either two right angles or together equal to two right angles.
Here we have given a straight line, standing on another straight line, and making two angles. And we have to prove that those angles are two right angles, or else are equal to two right angles.
Given. The line AB, standing on the line CD, and forming the two angles A B C, ABD.
Required. To prove either
(a) That angles ABC, ABD are two right angles, or
(b) That angles A B C, A B D are together equal to two right angles.
(a) Let us assume that angle A B C is equal to angle ABD.
Then we can form the following syllogism :-If a straight line standing on another straight line makes the two angles equal, each of them is a right angle.
The line AB standing on the line CD makes the angle ABC equal to the angle A B D. .. Each of the angles ABC, ABD is a right angle.
(b) Let us assume that angle A B C is not equal to angle A B D, as in this figure.
For the purpose of proof draw EB at right angles to CD.
In this figure we have now five angles.
Two angles CBA, ABD formed by the line AB, standing on CD:
Two angles CBE, EBD formed by the line EB, standing on CD:
And one angle E B A included between the lines BA.
(Try and keep these five angles separate in your mind during the following proof.)
If the same angle be added to equal angles the wholes are equal. (Axiom 2a.)
The angle DBE is equal to the angles DB A, ABE, add angle EBC to each of these equals;
(c) Then the two angles DBE, EBC are equal to the three angles DBA, ABE, EBC. If the same angle be added to equal angles the wholes are equal. (Axiom 2a.)
The angle CBA is equal to the angles CBE,
EBA, add angle ABD to each of these equals. (d) Then the two angles CBA, ABD are equal to the three angles CBE, EBA, ABD.