drawn from C and D meeting at B, the theorem teaches us that if the angles ABC, ABD are equal to two right angles, CB and BD are in the same straight line. The demonstration is a very simple example of the indirect method. Angles ABC, ABD being given equal to two right angles, we assert that A B -E D straight line). Then we have the angles ABC, ABE equal to two right angles according to the last proposition. Also angles ABC, ABD given equal to two right angles. From these pairs subtract the common angle ABC, and we have remaining angle ABE equal to angle ABD, the part to the whole, which is absurd. Hence it is evident that no other line but BD can be drawn in a straight line with CB. SUMMARY OF PROPOSITION XIV., THEOREM 7. If two straight lines meet a third at a point, making the adjacent angles equal to two right angles, these two lines shall be in the same straight line. Cons. If BD is not in the same straight line with CB, produce CB to E. Proof. Both pairs of angles = 2 rt. ≤ s, a (hyp.) ẞ (I, 13); subtract common angle, then ABD = ▲ ABE, which is absurd, C &c., &c. A E B D PROP. XV.-We pass to the third theorem of the group. If two straight lines cut one another, the vertical or opposite angles are equal. The vertical angles are those whose vertices or heads are exactly opposite. In some of the old editions of Euclid they are called the head angles. Let AB and CD be two straight lines cutting at the point E, then the vertical angles are equal, that is we have to show We have already said that the proof of this theorem mainly depends on the fact that all right angles are equal. Let us endeavour to apply this principle. A E C D First let us suppose that AB and CD cut at right angles, then it is obvious that there will B be four right angles in the figure and the vertical angles will be equal according to the postulate referred to (post. 4). Next, let the two straight lines cut at any other angle, then we shall not find any single right angles, but there will be pairs of angles, which by I. 13 will be together equal to two right angles. Let us examine the figure and see how many of these pairs we can obtain. CE falling on AB makes angles CEA, CEB= E Since these pairs of angles are each equal to two right angles, any pair must be equal to of the other pairs. In the first two any pairs of angles in our table we have a common angle CEA; take this away, and the remaining angle CEB = the remaining angle AED, and these are vertical angles. Again, in the second and third pairs we have a common angle AED; subtract this, and angle CEA = angle BED, the other pair of vertical angles. You will notice that the vertical angles might also be proved equal, by subtracting the common angle from the third and fourth pairs, and also from the first and fourth. Also, since each pair of angles is equal to two right angles the four angles made by one line cutting another are together equal to four right angles. Again, suppose that besides AB and CD in the figure we have any other lines, FG, HK, meeting or cutting at the point E, though each individual angle will be smaller you will see that the whole space is the same, and that all these angles must be together equal to four From these considerations we get two im right angles. portant corollaries. I. If two straight lines cut one another, the angles which they make at the point of section are together equal to four right angles. II. All the angles made by any number of straight lines meeting at one point are together equal to four right angles. From this group of theorems we learn the following facts about angles : If one straight line meets another the resulting angles are equal to two right angles. If one straight line cuts another, the angles formed are equal to four right angles, and also the vertical opposite angles are equal. If any number of straight lines meet in a point, the angles formed are equal to four right angles. SUMMARY OF PROPOSITION XV., THEOREM 8. If two straight lines cut one another, the vertical angles From these equals subtract common angle, and the remaining vertical angles are equal. In the same way the other pair may be shown equal. LECTURE VIII. PROPOSITIONS 16-19. PROPOSITION XVI. proves that if one side of a triangle be produced, the angle thus formed outside the triangle is greater than either of the angles which are opposite to it inside the triangle. Let ABC be a triangle of which the side BC is produced to D: then the exterior angle ACD shall be greater than either of the interior and opposite angles CBA, BAC. The construction made use of by Euclid for the proof of this theorem is exceedingly ingenious. We may imagine that it was arrived at in this way :— X According to the rule given in Lecture VI. for analysis, let us assume what we have to show, viz. that angle ACD is greater than angle BAC. Then we will make an angle ACX a part of ACD, which we will assume to be equal to angle BAC. To observe any results B E D from the equality of these angles we must, as in several other propositions, include them in equal triangles. How shall we do this? We have BAC = Z ACX. As the length of the arm of an angle does not affect the angle, we can make CX = BA; by bisecting AC in E (as in Proposition 12) we make the other arms of the angles equal; joining XE, EB, the extremities of the equal arms, we obtain the bases. The triangles thus constructed satisfy the conditions of I. 4, and are therefore equal. Next, can we retrace our steps, and construct these equal triangles, without assuming angle ECX? We can, as before, bisect AC in E, thus obtaining AE equal to EC. We can join EB, and produce it to F, making EF equal to EB. A F Do we know anything about the angle contained by these equal sides? Looking at the figure you will see that angle BEA is vertically opposite to angle FEC, and therefore these angles are equal (I. 15). Then, joining FC, we have constructed the triangles BAE, FEC equal = B D in all respects (1.4). Therefore the angle BAE (i.e. BAC) = angle ECF, a part of angle ACD. It is evident, then, that the angle ACD is greater than the angle BAC. It remains to show that angle ACD is greater than angle ABC. Produce AC to G. The angle BCG is evidently opposite the angle ACD, and is therefore equal to it (I. 15). Then, having bisected BC with the same construction as before, we can show that angle BCG is greater than angle ABC. Therefore, the angle ACD is also greater than angle ABC. Hence we have shown that the exterior angle ACD is greater than either of the interior and opposite angles. SUMMARY OF PROPOSITION XVI., THEOREM 9. The exterior angle of any triangle is greater than either of the interior and opposite angles. I. To show that ▲ ACD is greater than BAC. Cons.-Bisect AC in E. Join BE, produce to F, making EFEB. Join FC. Proof. - BAE = ^ ECF (I. 4). (AEB, FEC are vertical Zs, I. 15.) ../ ECF = BAC. ACD is greater than ▲ BAC. A II. To show that ▲ ACD is greater than ▲ ABC. Cons. Produce AC to G, bisect BC, &c. F D |