PROP. XIII. THEOREM. The angles which one straight line makes with another upon one side of it are either two right angles, or are together equal to two right angles. CASE I. When the angles are equal to each other. Let the straight line AB make with the straight line CD, on one side of it, the angles ABC and ABD = each other. Then it is to be proved that The angles ABC and ABD are two right angles. PROOF.-Because the angle ABC= the angle ABD (hyp.), therefore each of them is a right angle (def. 10). Therefore, it is proved, as required, that The angles ABC and ABD are two right angles. CASE II.-When the angles are not equal to each other. Let the straight line AB make with CD, on one side of it, the angles ABC and ABD not equal to each other. Then it is to be proved that The angles ABC and ABD are together two right CONSTRUCTION.-Draw BE at right angles to CD (I. 11). PROOF. The angles DBE and CBE two right angles (def. 10). Now, the angle EBC = the angles ABC and ABE, and, adding angle EBD to each of these, then the angles EBC and EBD the angles ABC, ABE, and EBD (ax. 2). 2 Again, the angle ABD = the angles ABE and EBD, and, adding angle ABC to each of these, then the angles ABD and ABC = the angles ABC, ABE, and EBD (ax. 2). But, as above, the angles EBC and EBD=the same angles ABC, ABE, and EBD; therefore the angles EBC and EBD = the angles ABD and ABC (ax. 1). But the angles EBC and EBD two right angles (cons.); therefore the angles ABD and ABC = two right angles (ax. 1). Therefore, it is proved, as required, that The angles ABC and ABD are together equal to two right angles. Wherefore, The angles which one straight line makes with another, &c. Q. E. D. 1 I.e. the double angle on the right side of the figure. PROP. XIV. THEOREM. If at a point in a straight line, two other straight lines, upon opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines shall be in one and the same straight line. At the point B, in the straight line AB, let the two straight lines BC and BD, on opposite sides of AB, make the adjacent angles ABC and ABD together two right angles. Then it is to be proved that = The two straight lines BC and BD are in one and the same straight line. CONSTRUCTION.-If BC and BD are not in the same straight line, then suppose that BC and BE are, i.e. that CBE is one straight line. PROOF.-Because CBE is a straight line, and AB is another line falling upon it in B, therefore the angles ABC and ABE together two right angles (I. 13). = = But the angles ABC and ABD together two right angles (hyp.), and therefore the angles ABC and ABD = the angles ABC and ABE (ax. 1). Take away the common angle ABC, and then the angle ABE = the angle ABD (ax. 3), i.e. a part the whole, which is absurd (ax. 9). Therefore the supposition that BC and BE are in the same straight line is false. = Similarly it can be proved that only BC and BD are in the same straight line. Therefore, it is proved, as required, that The two straight lines BC and BD are in one and the same straight line. Wherefore, If at a point in a straight line, &c. Q. E. D. PROP. XV. THEOREM. If two straight lines cut one another, then the vertical, or opposite, angles are equal to each other. Let the two straight lines AB and CD cut each other in the point E. Then it is to be proved that The angle AEC : the angle BED, and Similarly, The angle CEB = the angle AED. Ꭺ E B PROOF.-Because the angles AEC and AED = two right angles, and because also the angles AED and DEB=two right angles (I. 13); therefore the angles AEC and AED = the angles AED and DEB (ax. 1); take away the common angle AED from each, then the remaining angle AEC = the remaining angle DEB. Therefore, it is proved, as required, that If two straight lines cut one another, &c. COROLLARIES. Q. E. D. 1. If two straight lines cut one another, the four angles they make at the point where they cut, are together equal to four right angles. 2. If any number of straight lines cut one another, all the angles made by them where they cut, are together equal to four right angles. Exercises. 1. Prove in Prop. XV. that the angle CEB = the angle AED. 2. Prove each of the above Corollaries. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let the side BC be produced to D. Then it is to be proved that The exterior angle ACD shall be greater than either the interior opposite angles ABC or BAC.* CONSTRUCTION.-1. Bisect AC in E by the line BE (I. 10). 2. Produce BE to F (post. 2) making EF BE (I. 3), and join FC. PROOF.-Because in the triangles AEB and FEC, we have the sides AE, and EB, and their angle AEB, in the former = the sides CE and EF, and their angle FEC, in the latter, each to each (cons. and I. 15), therefore the angle BAE = the angle ECF (I. 4), i.e. the angle BAC the angle ACF (note 2 def. 15). But the angle ACD is greater than the angle ACF (ax. 9), i.e. the exterior angle ACD is greater than the interior opposite angle BAC. Similarly, If we bisect BC and produce AC to G, &c., as before, then the angle BCG would be proved to be greater than the angle ABC. |