are opposite at the vertex; and the other is the angle OIR, which is equal to the angle OFP; because the angles CIN and AFN are, in the query, supposed to be equal; and CIN and OIR are opposite angles at the vertex.

Q. But of what use is your proving that the triangle ORI is equal to the triangle OPF ?

A. It shows that since the triangle OPF is right-angular in P, the triangle OIR must be right-angular in R; for, in equal triangles, the equal angles are opposite to the equal sides. (query 6); consequently the two lines AB, CD, are both perpendicular to the same straight line PR, and therefore parallel to each other. Q. Supposing now

A

EFC equal, what relation would the lines AB,

CD, then bear to each other?

A. They would still be parallel.

Q. How do you prove this?

A. For if the angle AEF is equal to the angle EFD, the angles AEF and CFN will also be equal; because EFD and CFN are opposite angles at the vertex; and, in the same manner, if the angles BEF and EFC are equal, MEA and

EFC will also be equal; because MEA and BEF are opposite angles at the vertex. Therefore, in both cases, there will be two straight lines cut by a third line at equal angles, and consequently they will be parallel to each other.

Q. But there is one more case, and that is: If the two straight lines AB, CD (in our last figure), are cut by a third line MN, so as to make the sum of the two interior angles AEF and EFC equal to two right angles, how will the straight lines AB, CD, then be situated with regard to each other?

A. They will still be parallel to each other. For the sum of the two adjacent angles EFC and CFN is also equal to two right angles; and therefore, by taking from each of the equal sums the common angle EFC, the two remaining angles AEF and CFN must be equal (truth 7); and you have again the first case, viz: two straight lines cut by a third line at equal angles.

Q. Will you now state the different cases in which two straight lines must be parallel?

A. 1stly, When they are cut by a third line at equal angles.

2dly, When they are cut by a third line so as to make the alternate angles equal; and

3dly, When the sum of two interior angles, made by the intersection of a third line, are together equal to two right angles.

Q. Will you now repeat the demonstration of each separate case?

Supposing the two straight lines CD, EF, to be cut by a third line AM at unequal angles ABC, BHE, (Fig. I. and II.), or so as to have the alternate angles CBH and BHF, or DBH and BHE unequal; or in such a manner, that the sum of the two interior angles CBH and BHE (Fig. I.), or DBH and BHE (Fig. II.), is less than two right angles, what will then be the case with the two straight lines CD, EF?

A. They will, in every one of these cases, cut each other, if sufficiently extended.

Q. How can you prove this?

A. By drawing, through the point B, another line NP at equal angles with EF, and which will then also make the alternate angles NBH and

BHF, or PBH and BHE equal; and the sum of the two interior angles NBH and BHE equal to two right angles: the line NP will be parallel to the line EF; and then, if the line CD would not cut EF, it would also be parallel to it, and there would be, through the same point B, two lines NP, CD, drawn parallel to the same straight line EF, which cannot be (truth 10): Therefore the line CD is not parallel to EF; and therefore it must cut the line EF.

A. Yes. In the first place, the angle a is equal to the angle e; the angle c equal to the angle g; the angle b equal to the angle f; and the angle d equal to the angle h ;—2dly, the angles a, d, e, h, as well as the angles b, c, f, g, are respectively equal to one another; and finally, the sum of either c and e, or d and f, must make two right angles. For if either of these cases were not true the lines would not be parallel. (Query 10.)

QUERY XII.

From what you have learnt of the properties of parallel lines, what law can you discover respecting the distance they keep from each other? A. That parallel lines remain throughout equidistant.

Q. When do you call two lines equidistant? A. When all the perpendiculars, dropped from one line to the other, are equal.

Q. How can you prove, that the perpendicular lines OP, MI, RS &c. are all equal?

A. By joining MP the two triangles MPO, IMPI will have the side MP common; and the angle a is equal to the angle b; because a and b are alternate angles, formed by the two parallel lines MI, OP ( query 11 ); and the angle c is equal to the angle d; because these angles are formed in a similar manner by the parallel lines AB, CD therefore we have a side and the two adjacent angles in the triangle MPO, equal to a side and the two adjacent angles in the triangle MPI; consequently these two triangles must be equal; and the side OP opposite to the angle c, in the triangle MPO must be equal to the