GEOMETRY. 



Let the straight line CD make with AB upon one 

 side of it the two angles 

 ACD, BCD ; these two 

 angles are together equal 

 to two right angles. Since 

 AC and CB are in one line, 

 ACB is an angle of two 

 right angles, but the two angles ACD, BCD taken 

 together make up this angle ; hence the two angles 

 ACD, BCD are together equal to two right angles. 



Corollary. All the angles which can be made 

 round any point O are 

 together equal to four 

 right angles. For, pro- 

 duce the line AO 

 through O to E ; then 

 all the angles AOB, 

 BOC, COD, DOE are 

 together equal to two 

 right angles. Similarly, 

 the angles on the other side of AE are together 

 equal to two right angles, therefore all the angles 

 round O are together equal to four right angles. 



Conversely, if the two angles ACD, BCD 

 together make up two right angles, AC, CB are in 

 one line. For if CB 

 be not the continua- 

 tion of AC, let CE be 

 its continuation. Then 

 since ACE is a straight 



line, the two angles A c B 



ACD, ECD make two 



right angles; but ACD, BCD are also equal to 

 two right angles, therefore the one pair is equal 

 to the other ; and if ACD be taken from both, we 

 have the angle DCE equal to DCB, which is absurd, 

 therefore CB must be the continuation of AC. 



PROP. II. If two lines cut each other, the verti- 

 cal or opposite angles are equal. 



Let AB and CD cut each other in O ; it is 

 required to prove that the angle AOC is equal to 

 the angle BOD, or the angle COB equal to angle 

 AOD. Since CO makes 

 with AB the two angles 

 AOC, COB on the same 

 side of it, these are to- 

 gether equal to two right 

 angles. Similarly, the 

 two COB, BOD are equal 



to two right angles, therefore the one pair is equal 

 to the other ; and if COB be taken from both, we 

 have the angle AOC equal to the angle BOD. 



PROP. III. If one side of a triangle be pro- 

 duced: 



(i.) The exterior angle is equal to the two 

 interior remote angles. 



(2.) The exterior angle is greater than either of 

 the interior remote angles. 



(3.) The three interior angles are together equal 

 to two right angles. 



Let ABC be a triangle, having the side BC 

 produced to D : to prove 



(r.) That the two angles 

 ABC, BAG are together 

 equal to the exterior angle 

 ACD. 



(2.) That ACD is greater 

 than each separately. 



(3.) That the three in- 



terior angles of the triangle taken together make 

 up two right angles. 



(i.) Suppose the line BA or BE to have origin- 

 ally coincided with BD, and to have attained its 

 present position by revolving about B. Conceive 

 the line BE to be now broken as it were at A, 

 and that the part AE revolves about the point A, 

 into the position AF ; and let FA be produced to C. 

 The angle EBD or ABC measures the opening or 

 divergence between BE and its original position. 

 But the part AF diverges still farther from the 

 original position by the angle EAF. Therefore 

 AF diverges from BD, by the two angles ABC 

 and EAF, or BAG, which is equal to EAF (Prop. 

 II.). But the divergence or opening between FA 

 or FC and BD is the angle ACD ; therefore the 

 angle ACD is equal to the two angles ABC, BAG 

 taken together. 



(2.) Hence the exterior angle is greater than 

 either taken separately. 



(3.) Since the exterior angle ACD is equal to 

 the two interior angles ABC, BAG, if ACB be 

 added to each, we have the two ACD, ACB 

 together equal to the three ABC, BAG, ACB. 

 But the two ACD, ACB make up two right 

 angles, therefore the three ABC, BAG, ACB also 

 make two right angles. 



Cor. i. Any two angles of a triangle are to- 

 gether less than two right angles. 



Cor. 2. The interior angles of any rectilineal 

 figure are equal to twice as many right angles as 

 the figure has sides less four. 



Let ABCDE be a figure, say of five sides ; to 

 prove that the interior angles 

 taken together make up six 

 right angles. Take any point 

 O within the figure. Join O 

 with each of the angles. We 

 thus divide the figure into 

 as many triangles as the 

 figure has sides, that is, into 

 five, and the angles of each 

 triangle make up two right angles, hence ah 1 the 

 angles of the triangles are equal to ten right 

 angles ; but these angles include those round O, 

 which are four right angles, therefore the angles 

 of the figure make up six right angles. 



THE TRIANGLE. 



PROP. IV. The angles at the base of an isosceles 

 triangle are equal to each other. 



Let ABC be an isosceles triangle, having the 

 side AB equal to the side AC ; it is required to 

 prove that the angle ACB is equal 

 to the angle ABC. For if the tri- 

 angle ABC be lifted up, turned 

 round, and placed upon its old 

 position, so that A lies on A, and 

 so that AC lies on the former 

 position of AB, then AB shall lie 

 on the former position of AC ; B* 

 and as AC is equal to AB, the 

 point C shall lie on B, and the point B on C, and 

 the line CB shall lie on its former position, and 

 therefore the angle ACB shall be equal to the 

 angle ABC. 



Cor. i. If the equal sides be produced, the 

 angles on the other side are also equal. 



Cor. 2. If a triangle be equilateral, it is also 

 equiangular. 



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