= .. all the interior angles four right angles, and by post. 6 we know that neither of the pairs of interior angles can be less than two right angles; hence neither can be greater, or the four angles would be greater than four right angles. Therefore two interior angles taken on one side of the cutting line must be equal to two right angles. II. To show that the alternate angles are equal angles AGH + BGH = 2 rt. angles BGH + GHD = 2 rt. angles Take away the common angle BGH, and angle AGH = GHD, and they are alternate angles. III. As above. SUMMARY OF PROPOSITION XXIX., THEOREM 20. If two parallels are cut by a straight line : I. The alternate angles are equal. (a, Euclid's post.) Assume ▲ AGH greater than ▲ GHD. LAGH + BGH = 2 rt. angles (I. 13) :. ▲ GHD + BGH are less than 2 rt. angles. .. by post. 6, AB and CD will meet, which is impossible, for they are parallel, &c. (B, Playfair's post.) Assume AGH greater than ▲ GHD, make LGH = ▲ GHD (I. 23), and produce LG to M. LM is || to CD (I. 27), and cuts AB also || to CD, which is impossible, &c. II. The exterior angle is equal to the interior and opposite. LAGH GHD (pt. 1) = LAGH = 2 EGB (I. 15) ..the exterior 2 EGB= int. and opposite GHD. III. The two interior angles on one side of the line are equal to two right angles. ZAGH+ BGH 2 rt. angles (I. 13) ▲ AGH = GHD (pt. 1) 2 GHD + BGH = 2 rt. angles. LECTURE XIII. PROPOSITIONS 30-31.-ANALYSIS OF THE THEORY OF PARALLEL STRAIGHT LINES.-PROPOSITION 32 AND COROLLARIES. Two short and simple propositions, a theorem and a problem, conclude the subject begun in our last lecture. To show that straight lines which are parallel to the same straight line are parallel to one another. Let AB and CD be each of them parallel to EF, AB shall be parallel to CD. A C In the last proposition, three tests for parallel lines were given us; in order to apply them we must draw a line LM, cutting the given lines in GHK. Then we know that if AGH =▲ GHD, i.e. if the alternate angles are equal, AB is parallel to CD. Let us see if this is the case. E L B D F K M We know that ▲ AGH = ▲ HKF, because AB is parallel to EF; also we know that the exterior angle GHD is equal to the interior and opposite HKF, because CD is parallel to EF. Then by ax. 1, ▲ AGH = GHD, and it follows that AB is parallel to CD. SUMMARY OF PROPOSITION XXX., THEOREM 21. Straight lines which are parallel to the same straight line are parallel to each other. Cons.-Draw a line cutting the three given lines. ▲ GHD = HKF (ext. = .. AGH = GHD int. and opp. I. 28) Proposition 31 requires us to draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line. Let us assume that the problem is solved and that a line is drawn through A parallel to BC. What would be the result? Draw a line cutting the parallels, and then from I. 29 it follows that the alternate angles are equal, &c. E F These considerations furnish us with a hint for the construction, for by joining point A with any point D in BC we obtain a line AD, which must cut both parallels when drawn, and the angle ADC which it makes with BC will be one of the alternate angles, the other of which must evidently be at the point A. By I. 23 we are able to make an angle at this point equal to angle ADC, and producing EA one arm of the angle, through A to any point F, the construction is complete. For EF is parallel to BC, as the alternate angles are equal. B D SUMMARY OF PROPOSITION XXXI., PROBLEM 10. Through a given point to draw a straight line parallel to This problem concludes a group of five propositions dealing with the theory of parallel straight lines; in the three following theorems we apply the principles here learnt. Before proceeding to these we will sum up the results arrived at. First you will observe that the test for parallel lines is the equality of certain angles made by a line cutting the supposed parallels. Lines are parallel if : (a) The alternate angles are equal (I. 27). (3) The exterior angle is equal to the interior and opposite on the same side of the cutting line. (7) The two interior angles on the same side are together equal to two right angles (I. 28). The converse of these three theorems is true. And if any one of these facts is proved, the other two necessarily follow (I. 29). We have learnt also to draw a line parallel to another through any given point (I. 31). Also we may remark that parallel lines often give us a ready means of showing the equality of any angles which we may require for the demonstration of a theorem. : Angles can be shown to be equal, if: They form corresponding parts of equal triangles (I. 4, They are opposite equal sides in the same triangle. If two pairs of angles containing a common angle can I. 32. We pass to the investigation of Prop. XXXII., which may be considered as the completion of Props. XVI. and XVII. In these we learnt (a) that the exterior angle of a triangle is greater than either of the interior and opposite; we now show that it is equal to both. (B) That any two angles of a triangle are less than two right angles; we now prove that the three interior angles are together equal to two right angles. B C E D I. Let ABC be any triangle, and let BC be produced to D, we have to show that angle ACD is equal to the angles BAC+ ABC. We know (I. 17) that angle ACD is greater than BAC or ABC, and we know (I. 23) how to make an angle equal to another; let us then at the point C make an |