the sum of AGH, BGH is also equal to two right angles (Prop. I.); take away BGH, which is common to both, and there remains the angle GHD, equal to the angle AGH; and these are alternate angles; hence the lines AB, CD are parallel. PROPOSITION XXII.-THEOREM. 86. If a straight line intersects two parallel lines, it makes the alternate angles equal; also any exterior angle equal to the interior and opposite angle; and the two interior angles upon the same side together equal to two right angles. Let the straight line E F intersect the parallel lines AB, CD; the alternate angles AGH, GHD are equal; the exterior angle EG B is equal to the interior and opposite angle GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles. A K E L B C D H F For if the angle A G H is not equal to G H D, draw the straight line KL through the point G, making the angle KGH equal to GHD; then, since the alternate angles GHD, KGH are equal, KL is parallel to CD (Prop. XX.); but by hypothesis A B is also parallel to CD, so that through the same point, G, two straight lines are drawn parallel to CD, which is impossible (Art. 34, Ax. 12). Hence the angles A G H, GHD are not unequal ; that is, they are equal. Now, the angle EGB is equal to the angle A GH (Prop. IV.), and AGH has been shown to be equal to GHD; hence E G B is also equal to G H D. Again, add to each of these equals the angle BGH; then the sum of the angles EG B, BGH is equal to the sum of the angles B G H, G HD. But E G B, BG H are equal to two right angles (Prop. I.); hence BGH, GHD are also equal to two right angles. 87. Cor. If a line is perpendicular to one of two parallel lines, it is perpendicular to the other; thus EF (Art. 85), perpendicular to A B, is perpendicular to CD. PROPOSITION XXIII.-THEOREM. 88. If two straight lines intersect a third line, and make the two interior angles on the same side together less than two right angles, the two lines will meet on being produced. Let the two lines KL, CD make with EF the angles KGH, GHC, together less than two right angles; then KL and CD will meet on K being produced. For if they do not meet, they Care parallel (Art. 17). But they are not parallel; for then the sum E L D H F of the interior angles K GH, GHC would be equal to two right angles (Prop. XXII.); but by hypothesis it is less; therefore the lines KL, CD will meet on being produced. 89. Scholium. The two lines K L, CD, on being produced, must meet on the side of E F, on which are the two interior angles whose sum is less than two right angles. PROPOSITION XXIV.-THEOREM. 90. Straight lines which are parallel to the same line are parallel to each other. E Let the straight lines AB, CD be each parallel to the line EF; then are they parallel to each other. Draw GHI perpendicular to C EF. Then, since A B is parallel to EF, GI will be perpendicular A to AB (Prop. XXII. Cor.); and since CD is parallel to EF, GI will for a like reason be perpendicular to CD. Consequently AB and CD are perpendicular to the same straight line; hence they are parallel (Prop. XXI. Cor.). 91. Two parallel straight lines are everywhere equally distant from each other. Let AB, CD be two parallel straight lines. Through any two points in AB, as E and F, draw the straight lines EG, FH, perpendicular to A B. H C A These lines F will be equal to each other. G -D -B E For, if G F be joined, the angles GFE, F G H, considered in reference to the parallels A B, CD, will be alter nate interior angles, and therefore equal to each other (Prop. XXII.). Also, since the straight lines EG, FH are perpendicular to the same straight line AB, and consequently parallel (Prop. XXI. Cor.), the angles EGF, GFH, considered in reference to the parallels EG, FH, will be alternate interior angles, and therefore equal. Hence, the two triangles EFG, FGH, have a side and the two adjacent angles of the one equal to a side and the two adjacent angles of the other, each to each; therefore these triangles are equal (Prop. VI.); hence the side EG, which measures the distance of the parallels AB, CD, at the point E, is equal to the side FH, which measures the distance of the same parallels at the point F. Hence two parallels are everywhere equally distant. PROPOSITION XXVI.—THEOREM. 92. If two angles have their sides parallel, each to each, and lying in the same direction, the two angles are equal. Let ABC, DEF be two angles, which have the side AB parallel to DE, and BC parallel to EF; then these angles are equal. A D E F G B C H F E For produce DE, if necessary, till it meets BC in the point G. Then, since EF is parallel to G C, the angle DEF is equal to D G C (Prop. XXII.); and since D G is parallel to AB, the angle D G C is equal to ABC; hence the angle DEF is equal to ABC. 93. Scholium. This proposition is restricted to the case where the side EF lies in the same direction with BC, since if FE were produced toward H, the angles DEH, ABC would only be equal when they are right angles. 94. If any side of a triangle be produced, the exterior angle is equal to the sum of the two interior and opposite angles. Let ABC be a triangle, and let one of its sides, BC be produced towards D; then the exterior angle ACD is equal to the two interior and opposite B. angles, CA B, ABC. A E D For, draw EC parallel to the side A B; then, since AC meets the two parallels A B, EC, the alternate angles BAC, ACE are equal (Prop. XXII.). Again, since BD meets the two parallels AB, E C, the exterior angle ECD is equal to the interior and opposite angle ABC. But the angle ACE is equal to BAC; therefore, the whole exterior angle A CD is equal to the two interior and opposite angles CAB, ABC (Art. 34, Ax. 2). 95. In every triangle the sum of the three angles is equal to two right angles. Let ABC be any triangle; then will the sum of the angles ABC, BCA, CAB be equal to two right angles. For, let the side BC be pro- B duced towards D, making the A C -D exterior angle ACD; then the angle ACD is equal to CAB and ABC (Prop. XXVII.). To each of these equals add the angle A C B, and we shall have the sum of |