that Therefore, it is proved, as required in the second place, The straight lines AB and CD are parallel. Wherefore, If a straight line, &c. Q. E. D. Exercises. 1. Prove the above Proposition by taking the angles on the left of the cutting line EF, viz. the exterior angle EGA the interior and opposite angle GHC, and the two interior angles AGH and GHC. 2. If any angle BAC be bisected by a straight line AD, and any point E be taken in AD; prove that if straight lines EF and EG be drawn to AB and AC respectively, perpendicular to AD, then EF= EG. 3. If in a triangle ABC, with vertex C, the sides AC and BC be bisected at right angles by DF and EF respectively meeting in F; prove that FG drawn at right angles to the third side AB will bisect it in G. PROP. XXIX. THEOREM. If a straight line fall on two parallel straight lines it makes the alternate angles equal to one another; the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side taken together equal to two right angles. Let the straight line EF fall on the two parallel lines AB and CD. Then it is to be proved that 1. The alternate angle AGH = GHD. the alternate angle 2. The exterior angle EGB = the interior and opposite angle GHD. 3. The two interior angles BGH and GHD = two right angles. PROOF.-1. If the angle AGH is not equal to the angle GHD, then one of them must be greater than the other. Suppose that the angle AGH is greater than the angle GHD, and to each of them add the angle BGH; then the angles AGH and BGH are greater than the angles GHD and BGH (ax. 4). But the angles AGH and BGH = two right angles (I. 13); therefore the angles BGH and GHD must be less than two right angles; and therefore the straight lines AB and CD will meet if produced (ax. 12). But they never can meet when produced, because by hypothesis they are parallel (def. 35). Therefore the supposition that the angle AGH is greater than the angle GHD is erroneous. Similarly, the supposition that the angle AGH is less than the angle GHD might be shewn to be erroneous. Consequently, the angle AGH = the angle GHD. Therefore, it is proved, as required, that 1. The alternate angle AGH the alternate angle GHD. 2. Next: Because the angle EGB = the angle AGH (I. 15), and because, as it has just been proved, the angle AGH the angle GHD, therefore the angle EGB = the angle GHD (ax. 1). Therefore, it is proved, as required, that 2. The exterior angle EGB the interior and oppo 3. Further: Because, as we have just proved, the angle EGB the angle GHD, add to each of them the angle BGH, then the angles EGB and BGH the angles BGH and GHD (ax. 2). But the angles EGB and BGH = two right angles (I. 13), therefore the angles BGH and GHD =two right angles (ax. 1). Therefore, it is proved, as required, that 3. The two interior angles BGH and GHD = two right angles. Wherefore, If a straight line, &c. Q. E. D. Exercise. 1. Prove the above Proposition by taking 1. The alternate angles BGH and GHC; 2. The angles on the left of EF, viz. the exterior angle EGA, with its interior and opposite on the same side, GHC; and 3. The two interior angles AGH and GHC. Straight lines which are parallel to the same straight Let AB and CD be straight lines, parallel to the same straight line EF. Then it is to be proved that The straight lines AB and CD are parallel. CONSTRUCTION.-Let the straight line GHK cut the straight lines AB, EF, and CD, in the points G, H, and K, respectively. PROOF.-Because the straight lines AB and EF are parallel (hyp.), therefore the alternate angle AGH = the alternate angle GHF (I. 29). Next, because the straight lines EF and CD are parallel (hyp.), therefore the exterior angle GHF the interior and opposite angle HKD (I. 29). But it has been just proved that the angle AGH = the angle GHF; therefore the angle AGH the angle HKD, i.e. the angle AGK = the angle GKD (note 2 def. 15); and these are alternate angles. Therefore, it is proved, as required, that The straight lines AB and CD are parallel (I. 27). Wherefore, Straight lines, &c. Q. E. D. PROP. XXXI. PROBLEM. 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. It is required to draw a straight line through A, which shall be parallel to BC. CONSTRUCTION.-1. In the straight line BC, take any point, D, and join AD. 2. At the point A in the straight line AD make the angle EAD the angle ADC (I. 23), and produce the straight line EA to F. Then it is to be proved that The straight line EF drawn through the point A, is parallel to the straight line BC. PROOF. Because the straight line AD falling upon the two straight lines EF and BC makes the alternate angle EAD the alternate angle ADC (cons.), therefore EF is parallel to BC (I. 27). = Therefore, it is proved, as required, that The straight line EF drawn through the point A is parallel to the straight line BC. Q. E. F. Exercise. If from the extremities of two equal and parallel straight lines AB and CD, straight lines AD and BC are drawn, joining their extremities and intersecting in E; prove that AEED and that BE = EC. D |