GHD. [Ax. 1.] Add to each of these the angle BGH. Therefore the angles EGB, BGH are equal to the angles BGH, GHD. [Ax. 2.] But the angles EGB, BGH are together equal to two right angles. [I. 13.] Therefore the angles BGH, GHD are together equal to two right angles. [Ax. 1.] Therefore, if a straight line, &c. Q.E.D. Ex. 1. The perpendiculars to two parallel lines are themselves parallel. 2. If the angle ABC be equal to PQR, and AB be parallel to PQ, then if BC and QR lie on the same side of AB and PQ respectively BC shall also be parallel to QR. 3. Any straight line drawn through the middle point of the diagonal of a parallelogram will cut off equal parts from its sides. PROPOSITION 30. THEOREM. Straight lines which are parallel to the same straight line are parallel to each other. 1. Let AB, CD be each of them parallel to EF: 2. Let the straight line GHK cut AB, EF, CD. to the angle GKD. [I. 29.] And it was proved that the angle AGK is equal to the angle GHF. Therefore the angle AGK is equal to the angle GKD; [Ax. 1.] and they are alternate angles; therefore AB is parallel to CD. [I. 27.] Therefore, straight lines, &c. Q.E.D. PROPOSITION 31. PROBLEM. To draw a straight line through a given point parallel to a given straight line. 1. Let A be the given point, and BC the given straight line: it is required to draw a straight line through the point A parallel to the straight line BC. In BC take any point D, and E. 2. join AD; at the point A in the straight line AD, make the angle B DAE equal to the angle ADC; [I. 23.] and produce the straight line EA to F. EF shall be parallel to BC. Because the straight line AD, which meets the 3. two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another. [Const.] EF is parallel to BC. [I. 27.] Therefore the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Q.E.F. PROPOSITION 32. THEOREM. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides 1. BC be produced to D: the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, namely, ABC, BCA, CAB shall be equal to two right angles. Through the point C draw CE parallel to AB. 2. [I. 31], is parallel to CE, and BD falls on them, the exterior ECD is equal to the interior and opposite angle ABC. [I. 29.] But the angle ACE was shown to be equal to the angle BAC; therefore, the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC. [Ax. 2.] To each of these equals add the angle ACB; therefore, the angles ACD, ACB are equal to the three angles CBA, BAC, ACB. [Ax. 2.] But the angles ACD, ACB are equal to two right angles [L. 13.]; therefore, also the angles CBA, BAC, ACB are equal to two right angles. [Ax. 1.] Therefore, if a side of any triangle, &c. Q.E.D. COROLLARY I. All the exterior angles of any rectilineal figure are together equal to four right angles. Let ABCD be any rectilineal figure having the 1. exterior angles A, B, C, D, these shall be together equal to four right angles. fore the exterior angles are equal to the angles a, b, c, and d. But these angles are equal to four right angles. [Cor. I. 15.] Therefore the exterior angles are equal to four right angles. Therefore, &c. COROLLARY 2. All the interior angles of any rectilineal figure together with four right angles are equal to twice as many right angles as the figure has sides. 3. For any interior angle ABD, together with its adjacent exterior angle DBC is equal to two right angles. Therefore, all the interior angles and all the exterior angles are equal to twice as many right angles as the figures has sides. But all the exterior angles are by the preceding corollary equal to four right angles. Α' B Therefore, all the interior angles and four right angles are equal to twice as many right angles as the figure has sides. N.B. In the case of a regular figure, as all its angles are equal, the value of an exterior angle will be found by dividing four right angles by the number of sides of the figure; and if the quotient be subtracted from two right angles the result will be the value of an interior angle. This proposition and its corollaries show what is the value of the interior and exterior angles of rectilineal figures. The second corollary is true of all rectilineal figures. The first is only true of convex figures which have no re-entrant angle, ie,, no angle which is greater than two right angles. Let ABCDE be a concave figure having the re-entrant angle EDC (which is dotted) greater than two right angles by the excess EDF. Then by drawing lines from B parallel to the sides it will be seen, as in Cor. 1, that the exterior angles are greater EXERCISES. 1. If any angle of a triangle be equal to the sum of the other two, it must be a right angle; but if it be greater than the sum of the other two it must be an obtuse angle, and if less an acute angle. 2. If the side CB of an equilateral triangle ABC be pro- 4. The middle point of the hypotenuse of a right angled PROPOSITION 33. THEOREM. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel. Let AB and CD be equal and parallel straight 1. lines, and let them be joined towards the same parts by the straight lines AC and BD: AC and BD shall be equal and parallel. 2. Join BC. |