156 



THE POPULAR EDUCATOE. 



Delightful in his manners inflexible in his principles and gene- 

 rous in his affections, he had all that could charm in society, or attach 

 in private. 



The joys of life in hurried exile go till hope's fair smile, and 

 beauty's ray of light, are shrouded in the griefs and storms of night. 



Bay after day prepares the funeral shroud ; the world is grey with 

 age : the striking hour is but an echo of death's summons loud the 

 jarring of the dark grave's prison door. Into its deep abyss devour- 

 ing all kings and the friends of kings alike must fall. 



She made an effort to put on something like mourning for her son ; 

 and nothing could be more touching than this struggle between pious 

 affection and utter poverty : a black ribbon or so a faded black hand- 

 kerchief, and one or two more such humble attempts to express by 

 outward signs that grief that passeth show. 



LESSONS IN GEOMETRY. V. 



SIMPLE GEOMETEICAL THEOEEMS. 



BEFOKE entering on the consideration of problems in geometry 

 which will be found to be practically useful to all who are 

 engaged in any mechanical art, it will be necessary for the 

 learner to become acquainted with a few simple statements or 

 facts in geometry, the truth of which is so clear and plain that 

 they require but little, if any explanation. These are called 

 theorems, or self-evident propositions, from the Greek eeupr)/j.a 

 (the-o-re'-ma), literally a sight, or something which can be 

 seen, in contradistinction to problems, or propositions which 

 require something to be done in order to effect their solution. 

 The word "problem" is derived from the Greek irpo/3A.rjjua 

 (pro-ble'-ma), which is derived in its turn from irpo (pro) before, 

 and $aAAo> (bal-lo) to cast >r throw, while the word "propo- 

 sition " is derived from the Latin pro, before, and pono, to 

 place. Hence the meaning of the words " problem " and " pro- 

 position " is precisely the same, namely, something that is 

 placed before you to be done or solved. 



1. When one straight line intersects another straight line, the 

 vertical or opposite angles are equal to one another. 



Let the straight line A B intersect the straight line c D in the 

 point E. Now, by the intersection of 

 these two straight lines, four angles v \ 



are formed, namely, c E A, A E D, ^.^ 



DEB, and EEC. Of these the ver- JL"~~ E""--^. B 

 tical or opposite angles are equal, ^^"-^ 



namely, c E A to DEB, and A E D . 

 to c E B. Fig. 1. 



The truth of this may be shown in 



a very simple and practical manner by copying the figure on a 

 piece of paper, and then cutting out the angles and placing 

 them on each other, the greater on the greater and the less on 

 the less. This mode of proof will frequently be found useful in 

 similar cases. 



Opposite angles are also called vertical angles, because the 

 top or vertex of each angle is directly opposite to the vertex of 

 the other. 



2. When a straight line intersects two parallel straight lines, 

 the alternate angles are equal. 



Let the straight line E F intersect the parallel straight lines 



A B, c D, in the points G H. The angles 



A G H, G H D are alternate angles, and 



\ are equal to one another, and the angles 



* \- B c H G, H G B are also alternate and 



\ equal. 



c \ D There are eight angles formed by 



\ the intersection of the straight 'lines 



p A B, c D, E F, in Fig. 2. Of these the 



Fig. 2. reader will find that there are two sets 



of four angles that are equal to one 



another namely, AGE = BGH = GHC = DHF, and E G B 

 = AGH = GHD = CHF. Let him demonstrate the truth of 

 this practically by drawing the figure on paper, cutting out 

 one of the greater angles and one of the less, and placing them 

 on the remaining angles in each set of four. 



3. The adjacent angles which are formed when one straight 

 line stands on another straight line, are together equal to tivo 

 right angles. 



In Fig. 3 ' the adjacent angles ABC, A B D, which are 

 formed by the straight line A B standing on the straight line 

 c D, are equal to two right angles. The truth of this is 

 evident? when we consider that each of the angles C B K, 



Fig. 4. 



DEE is a right angle, the straight line B E being at right 



angles to the straight line c D, and making the adjacent 



angles D B E, EEC equal to one ano- ^ 



ther. The pupil will remember that the 



measure of an angle is the extent of the 



opening of the lines or legs of which the 



angle is formed. Thus, the sum of the 



openings of the two angles ABC, A B D, 



or the sum of the openings of the three D 



angles c B A, A B E, E B D is equal to the sum Fig. 3. 



of the openings of the angles c B E, E B D. 



Thus we learn that if any number of straight lines meet in a 

 point in another straight line on one side of it, the sum of the 

 angles which they make with this straight line and with each 

 other are equal to two right angles ; and if any number of 

 straight lines meet in the same point on the other side of it, the 

 angles thus made are also equal to two right angles. Hence 

 the angles made by any number of lines meeting together in the 

 same point are together equal to four right angles. 



As a familiar illustration of this, the spokes of a wheel may 

 be taken, which radiate from the nave as a common centre. If 

 a chalk" line were drawn down the middle of each spoke, these 

 lines would meet in the centre of the nave, and the angles 

 formed by these lines at their point of meeting would be equal 

 to four right angles. 



4. Any awgle drawn in a semicircle is a right angle. 



An angle drawn in a semicircle is one which has its top or 

 vertex in the arc, while its legs pass through the extremities of 

 the diameter at its points of contact with 

 the arc. Thus, the angle A c B in the 

 semicircle A c B is a right angle. The 

 truth of this may be shown by cutting 

 out a right-angled triangle and applying 

 it to a semicircle. If large enough, it 

 will be found that the legs of the right 

 angle will pass through the ends of the diameter of the semi- 

 circle, no matter at what point in the arc of the semicircle the 

 vertex of the right angle may be placed. 



5. The greatest side of every triangle is opposite the greatest 

 angle. 



In the triangle A B c in Fig. 5, of the three angles A B C, B c A, 

 c A B A B c is manifestly the greatest ; B ^ 



while of the three straight lines 

 A B, B c, c A, which form its sides, 

 A c is the greatest. A c, the greatest 

 side, is opposite the greatest angle 

 ABC; or, in other words, A c, the 

 greatest side, subtends the greatest 

 angle ABC. 



A moment's reflection will show that the greatest angle of 

 any triangle must have the greatest opening between the lines 

 of which it is formed, and that the line which is opposite to or 

 subtends the greatest opening, must of necessity be greatest of 

 the three lines which subtend the three openings of the angles 

 of the triangle. 



6. If one side of a triangle be produced, the outer or exterior angle 

 is equal to the two interior and opposite angles of the triangle. 



In the figure that accompanies the preceding theorem let the 

 side A c of the triangle A B c be produced to D. The outer or 

 exterior angle B c D is equal to the two interior and opposite 

 angles c B A, B A c. For if at the point c in the straight line 

 A D the straight line c E be drawn parallel to A B, then the 

 alternate angles E c B, c B A are equal to one another, and by 

 Theorem 2, the angle D c E is equal to the angle CAB; but the 

 angles D c E, E c B together make up the angle D c B, which is 

 therefore equal to the angles c B A, B A c. 



7. The three interior angles of every triangle are together equal 

 to hvo right angles. 



In Fig. 5 the angle BCD has been shown to be equal to the 

 angles C B A, B A c ; to each of these equals add the angle B c A. 

 Now, by Theorem 3 the angles D c B, B c A are equal to two 

 right angles, and C B A, B A C, A C B, the three interior angles 

 of the triangle ABC, which are equal to these two angles, must 

 therefore be equal to two right angles. 



i-EOBLEMS IN PEACTICAL GEOMETEY. 



PROBLEM: A. To bisect a given straight line that is, to divide 

 U into two equal parts. 



Fig. 5. 



