Because the angle EGB is equal to the angle GHD, and the angle EGB equal (I. 15.) to the angle AGH, 1. The angle AGH is equal to the angle GHD, and they are alternate angles; therefore (I. 27.) 2. AB is parallel to CD. Again, because the angles BGH, GHD, are equal (Hyp.) to two right angles, and that AGH, BGH, are also equal (I. 13.) to two right angles, 1. The angles AGH, BGH, are equal to the angles BGH, GHD. Take away the common angle BGH; therefore 2. The remaining angle AGH is equal to the remaining angle GHD, and they are alternate angles ; therefore (I. 27.) 3. AB is parallel to CD. Wherefore if a straight line, &c. Q.E.D. PROP. XXIX.-THEOREM. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD, are equal to one another; and the exterior angle EĞB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD, upon the same side, are together equal to two right angles. For if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater; and because the angle AGH is greater than the angle GHĎ, add to each of them the angle BGH; therefore 1. The angles AGH, BGH, are greater than the angles BGH, GHD; but the angles AGH, BGH, are equal (I. 13,) to two right angles ; therefore 2. The angles BGH, GHD, are less than two right angles; but those straight lines which, with another straight line falling upon them, make the interior angles on the same side less than two right angles, will meet (Ax. 12.) together if continually produced; therefore 3. The straight lines AB, CD, shall meet, if produced far enough ; but they never meet, since they are parallel by the hypothesis ; therefore 4. The angle AGH is not unequal to the angle GHD, that is, it is equal to it; but (I. 15.) the angle AGH is equal to the angle EGB; therefore likewise 5. EGB is equal to GHD; add to each of these the angle BGH; therefore 6. The angles EGB, BGH, are equal to the angles BGH, GHD; but EGB, BGH, are equal (I. 13.) to two right angles ; therefore also 7. BGH, GHD, are equal to two right angles. Wherefore, if a straight line, &c. Q.E.D. PROP. XXX.-THEOREM. Straight lines which are parallel to the same straight line are parallel to one another. Let AB, CD, be each of them parallel to EF; AB is also parallel to CD. . Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, therefore (I. 29.) 1. The angle AGH is equal to the angle GHF. Again, because the straight line GHK cuts the parallel straight lines EF, CD, (I. 29.) 2. The angle GHF is equal to the angle GKD; and it was shown that the angle AGK is equal to the angle GHF; therefore also 3. AGK is equal to GKD; and they are alternate angles; therefore (I. 27.) 4. AB is parallel to CD. 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 the point 4, parallel to the straight line BC. In BC take any point D, and join 4D; and at the point 4, in the straight line AD, make (I. 23.) the angle DAE equal to the angle ADC'; and produce the straight line FA to F; EF' is parallel to BC. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC, equal to one another, (I. 27.) EF is parallel to BC. Therefore the straight line EAF is drawn through the given point A: parallel to the given straight line BC. Which was to be done. PROP. XXXII.- -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 BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangles, viz. ABC, BCA, CAB, are together equal to two right angles. Through the point C draw CE parallel (I. 31.) to the straight line AB; and because AB is parallel to CÈ, and AC meets them, (I. 29.) 1. The alternate angles BAC, ACE, are equal. Again, because AB is parallel to CE, and BD falls upon them, 2. The exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to the angle BAC; therefore (Ax. 2.) 3. The whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to each of these equals add the angle ACB; and 4. The angles ACD, ACB, are equal to the three angles CBA, BAC, ACB; but the angles 4CD, ACB, are equal (I. 13.) to two right angles; therefore also 5. The angles CBA, BAC, ACB, are equal to two right angles. Wherefore, if a side of any triangle, &c. Q.E.D. Cor. 1.--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. E A For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And by the preceding proposition, ali the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is, (I. 15. Cor. 2.) together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. COR. 2.- All the exterior angles of any rectilineal figure are together equal to four right angles. Because every interior angle ABC, with its adjacent exterior ABD, is equal (I. 13.) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles. PROP. XXXIII.-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, CD, be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD, are also equal and parallel. A B с Join BC; and because AB is parallel to CD, and BC meets them, (I. 29.) 1. The alternate angles ABC, BCD, are equal. and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC, are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore (I. 4.) 2. The base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, each to each, to which the equal sides are opposite ; therefore 3. The angle ACB is equal to the angle CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD, equal to one another, (I. 27.) 4. 4C is parallel to BD; and it was shown to be equal to it. Therefore, straight lines, &c. Q.E.D. PROP. XXXIV. -THEOREM. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. N.B.-A parallelogram is a four-sided figure, of which the opposite sides are parallel ; and the diameter is the straight line joining two of its opposite angles. Let ACDB be a parallelogram, of which BC is a diameter ; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it. A B D |