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OF EQUALITY AND SIMILARITY OF TRIANGLES.
OF THE EQUALITY OF TRIANGLES.
Preliminary Remark. There are three kinds of equality to be considered in triangles, viz: equality of area without reference to the shape; equality of shape without reference to the area, or similarity; and equality of both shape and area, or coincidence. All questions asked in this section will refer only to the last two kinds of equality; and those in the first part, only to the coincidence of triangles.
If in two triangles, two sides of the one are equal to two sides of the other, each to each, and the angles which are included by them also equal to one another, what relation will these two triangles bear to each other?
Ans. They will be equal to each other in all their parts, that is, they will coincide with each other throughout.
Show me that this must be the case with the two triangles, ABC and ab c, in which we
will suppose the side AB = ab, AC = ac, and the angle at A equal to the angle at a.
A. By placing the line a cupon its equal, AC, the angle at a will coincide with the angle at A, because these two angles are equal; and the line ab will fall upon the line AB; and as a b = AB the point b will fall upon B; that is, the three points of the triangle ab c will fall upon the three points of the triangle ABC, thus: The point a upon A,
C; consequently these two triangles will coincide.
Q. What remark can you make with respect to the sides and angles of triangles which coincide with each other ?
A. That the equal sides, cb, CB, are opposite to the equal angles, at a and A.
If one side and the two adjacent angles in one triangle, are equal to one side and the two adjacent angles in another triangle, each to each, what relation will the two triangles bear to each other?
A. They will be equal, and the angles opposite to
the equal sides will also be equal, as has been proved in the 1st Section. (Query 6.)
What remark can you make with respect to the two angles at the basis of an isosceles triangle ? A. They are equal to each other. Q. How can you prove
A. Suppose we had two equal isosceles triangles, ABC and abc, or B as it were, another impression, a b c, of the triangle ABC, that is, The side ab = AB,
ac = AC,
bc=BC. The angle at a= angle at A,
C. Then the sides, AB, AC, ab, a c, being all equal to one another, and the angle at a remaining the same, whichever way we place it, the whole of the two triangles, ab cand ABC, will still coincide, when ab c is placed upon ABC in such a manner that a c will fall upon AB, and ab upon AC (for you will still have two sides and the angle which is included by them in the one, equal to two sides and the angle which is included by them in the other); and therefore the angle at c
must coincide with the angle at B; and as the angle at c is only as it were, another impression of the angle at C, B and C must also coincide; that is, the two angles, B and C, at the basis of the isosceles triangle ABC, must be equal : and the same can be proved of the two angles at the basis of every other isosceles triangle.
If the three sides of one triangle are equal to the three sides of another, each to each, what relation will the two triangles bear to each other?
A. They will coincide throughout; that is, their angles will also be equal, each to each. Q. How can you prove
с this, for instance, of the two triangles ABC and abc, in which we will suppose A
B the side AB=ab,
BC=bc? That you may easier find out your demonstration, I have placed the two triangles, as you see, along side of each other, with their bases, AB and ab, together, and have joined their opposite vertices, C and c, by the straight line Cc. What can you now observe with regard to the two triangles AC c and BC c?