that point, you draw a perpendicular to any of the lines, for instance the perpendicular MN to the line OP, all the angles, a, b, c, d, e, f, taken together, occupy the same space which is occupied by the four right angles, formed by the intersection of the two perpendiculars MN, OP.


If a triangle has one side, and the two adjacent angles, equal to one side and the two adjacent angles of another triangle, what relation must these triangles bear to each other ?

A. They must be equal.

Q. Supposing in this diagram the side a b equal to AB; the angle at a equal to the angle at A, and the angle at b equal to the angle at B; how can you prove that the triangle abc is equal to the triangle ABC ?



B a A. By applying a b upon its equal AB, the side a c will fall upon AC, and b c upon BC, because the angles a and A, b and B, are respectively equal; and as the sides a c, b c, take the same direction as the sides AC, BC, they must also meet in the same point in which the sides AC, BC, meet; that is, the point c will fall up

on C; and the triangles will coincide throughout.

Q. What relation do you here discover between the equal sides and angles ?

A. That the equal angles c, C, are opposite to the equal sides a b, AB respectively.


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If two straight lines are both perpendicular to another straight line, what relation must they bear to each other ?

A. They must be parallel.
Q. Let us sup-

pose the two lines
AB, CD, to be both

A: perpendicular to a third line, GH; how can you convince

me that AB and CD
are parallel ?
A. Because, if

you extend AB and
CD in the directions
BE, DF, making BE

N and DF equal to BA and DC respectively, everything will be equal on both sides of the line GH. Now if the lines AB, CD, are not parallel, they must either be converging or diverging. If they are converging, AB and CD will, when sufficiently extended, cut each other somewhere in M;

but then the same must take place with the lines BE, DF, on the other side of the line GH, which will cut each other somewhere in N; and there would be two straight lines cutting each other in two points, which is impossible. If the lines AB, CD, were diverging, BE, DF, would be the same; but it is equally absurd to suppose two straight lines to diverge in two directions: consequently, the two straight lines, AB, DC, can neither be converging nor diverging, and therefore they must be parallel.

Q. Can two straight lines which cut each other, be perpendicular to the same straight line ?

A: No.
Q. Why not?

A. Because, if they are both perpendicular to a third line, I have just proved that they must be parallel ; and if they are parallel, they cannot cut each other.

Q. From a point without a straight line, how many perpendiculars can there be drawn to that same straight line ?

A. Only one.
Q. Why can there not more be drawn?

A. Because I have proved that two perpendiculars to the same straight line must be parallel to each other; and two lines, parallel to each other, cannot be drawn from one and the same point.


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If one of two

A parallel lines, for instance the line AB, is perpendicuM

-N lar to a third line

B D MN, does it follow that the other parallel line CD, must also be perpendicular to it ?

A. Yes. Because I have proved (query 7) that two lines, which are both perpendicular to a third line, must be parallel; and this is only the reverse of it.

Q. Your answer is not quite satisfactory. For some might imagine it possible that another line, DE, likewise parallel to AB, could, at the same time, be perpendicular to MN.

A. But then there would be two straight lines, drawn from the same point D, parallel to the same straight line AB, which cannot be. (Truth 10.)

Will you now repeat your whole demonstration, and show that if one of two parallel lines is perpendicular to a straight line, the other line must also be perpendicular to it?

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If a straight line

A MN, cuts two other straight lines

at equal angles ; that is, so as to make the


-N angles CIN and

R AFN equal ; what relation exists be

D tween these two lines ?

A. They are parallel to each other.

Q. How can you prove it by this diagram? The line IF is bisected in 0, and, from that point 0, a perpendicular OP is dropped to the line AB, and afterwards extended until, in the point R, it strikes the line CD.

A. I should first observe that the triangles OPF and ORI are equal; because the triangle OPF has a side and two adjacent angles equal to a side and two adjacent angles of the triangle ORI, each to each. (Query 6.)

Q. Which is that side, and which are the two adjacent angles ?

A. The side OI, which is equal to OF; because the point o bisects the line IF. One of the two adjacent angles is the angle IOR, which is equal to the angle FOP; because these angles

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