8. If two intersecting right lines be respectively parallel to two others, the angle between the former is equal to the angle between the latter. For if AB, AC be respectively parallel to DE, DF, and if AC, DE meet in G, the angles A, D are each equal to G [xxix.]. PROP. XXX.—Theorem. If two right lines (AB, CD) be parallel to the same right line (EF), they are parallel to one another. Dem.—Draw any secant GHK. Then since AB and EF are parallel, the angle AGH is equal to GHF [xxix.]. In like manner the angle GHF is equal to HKD [xxix.]. Therefore the angle AGK is equal to the angle GKD (Axiom i.). Hence [xxvii.] AB is parallel to CD. PROP. XXXI.—Problem. Through a given point (C) to draw a right line parallel to a given right line. Sol.—Take any point D in AB. Join CD (Post. i.), and make the angle DCF equal to the angle ADC [xxiii.]. The line CE is parallel to AB [xxvii.]. Exercises. 1. Given the altitude of a triangle and the base angles, construct it. 2. From a given point draw to a given line a line making with it an angle equal to a given angle. Show that there will be two solutions. 3. Prove the following construction for trisecting a given line AB:—On AB describe an equilateral △ ABC. Bisect the angles A, B by the lines AD, BD, meeting in D; through D draw parallels to AC, BC, meeting AB in E, F: E, F are the points of trisection of AB. 4. Inscribe a square in a given equilateral triangle, having its base on a given side of the triangle. 5. Draw a line parallel to the base of a triangle so that it may be—1. equal to the intercept it makes on one of the sides from the extremity of the base; 2. equal to the sum of the two intercepts on the sides from the extremities of the base; 3. equal to their difference. Show that there are two solutions in each case. 6. Through two given points in two parallel lines draw two lines forming a lozenge with the given parallels. 7. Between two lines given in position place a line of given length which shall be parallel to a given line. Show that there are two solutions. PROP. XXXII.—Theorem. If any side (AB) of a triangle (ABC) be produced (to D), the external angle (CBD) is equal to the sum of the two internal non-adjacent angles (A, C), and the sum of the three internal angles is equal to two right angles. Dem.—Draw BE parallel to AC [xxxi.]. Now since BC intersects the parallels BE, AC, the alternate angles EBC, ACB are equal [xxix.]. Again, since AB intersects the parallels BE, AC, the angle EBD is equal to BAC [xxix.]; hence the whole angle CBD is equal to the sum of the two angles ACB, BAC: to each of these add the angle ABC and we have the sum of CBD, ABC equal to the sum of the three angles ACB, BAC, ABC: but the sum of CBD, ABC is two right angles [xiii.]; hence the sum of the three angles ACB, BAC, ABC is two right angles. Cor. 1.—If a right-angled triangle be isosceles, each base angle is half a right angle.