Therefore, the planes MN, PQ cannot meet on being produced; hence they are parallel to each other. PROPOSITION XIII.-THEOREM. 420. If two parallel planes are cut by a third plane, the tiro intersections are parallel. Let the two parallel planes MN and PQ be cut by the plane EFGH, and let their intersections with it be EF, GH; then EF is parallel to G H. For, if the lines E F, GH, lying in the same plane, were not parallel, they would meet each P MA F E H G N other on being produced; therefore the planes MN, PQ, in which those lines are situated, would also meet, which is impossible, since these planes are parallel. PROPOSITION XIV.-THEOREM. 421. A straight line which is perpendicular to one of two parallel planes, is also perpendicular to the other plane. Let MN, PQ be two parallel planes, and AB a straight line perpendicular to the plane MN; then A B is also perpendicular to the plane PQ. P M A D N B Draw any line, BC, in the plane PQ; and through the lines A B, B C, conceive a plane, ABC, to pass, intersecting the plane M N in AD; the intersection AD will be parallel to BC (Prop. XIII.). But the line A B, being perpendicular to the plane M N, is perpendicular to the straight line AD; consequently it will be perpendicular to its parallel BC (Prop. XXII. Cor., Bk. I.). Hence the line A B, being perpendicular to any line, BC, drawn through its foot in the plane PQ, is consequently perpendicular to the plane PQ (Art. 388). PROPOSITION XV.—THEOREM. 422. Parallel straight lines included between two parallel planes are equal. Let EF, GH be two parallel straight planes, included between two parallel planes, MN, PQ ; then EF and GH are equal. For, through the parallels EF, GH conceive the plane E F G H to pass, intersecting the parallel planes in EG, FH. The inter M E P H F sections EG, FII are parallel to each other (Prop. XIII.); and EF, GH are also parallel; consequently the figure E F G H is a parallelogram; hence EF is equal to GH (Prop. XXXI. Bk. I.). 423. Cor. Two parallel planes are everywhere equidistant. For, if EF, GH are perpendicular to the two planes MN, PQ, they will be parallel to each other (Prop. X. Cor. 1); and consequently equal. PROPOSITION XVI.-THEOREM. 424. If two angles not in the same plane have their sides parallel and lying in the same direction, these angles will be equal, and their planes will be parallel. Let BAC, EDF be two triangles, lying in different planes, MN and PQ, having their sides. parallel and lying in the same direction; then the angles BAC, EDF will be equal, and their planes, MN, PQ, be parallel. M H G D E and parallel to BE. For a similar reason, CF is equal and parallel to AD; hence, also, B E is equal and parallel to CF; hence the figure BCFE is a parallelogram, and the side BC is equal and parallel to E F; therefore the triangles BA C, EDF have their sides equal, each to each; hence the angle B A C is equal to the angle ED F. Again, the plane BAC is parallel to the plane EDF. For, if not, suppose a plane to pass through the point A, parallel to ED F, meeting the lines BE, CF, in points different from B and C, for instance G and H. Then the three lines GE, AD, HF will be equal (Prop. XV.). But the three lines BE, AD, CF are already known to be equal; hence BE is equal to G E, and HF is equal to CF, which is absurd; hence the plane B A C is parallel to the plane EDF. 425. Cor. If two parallel planes M N, PQ, are met by two other planes, A BED, ACFD, the angles BA C, EDF, formed by the intersections of the parallel planes, are equal; for the intersection A B is parallel to ED, and AC to DF (Prop. XIII.); therefore the angle B A C is equal to the angle ED F. PROPOSITION XVII. THEOREM. 426. If three straight lines not in the same plane are equal and parallel, the triangles formed by joining the extremities of these lines will be equal, and their planes will be parallel. Let BE, AD, CF be three equal and parallel straight lines, not in the same plane, and let B A C, E D F be two hence, the side AB is equal and parallel to DE (Prop. XXXIII. Bk. I.). For a like reason, the sides BC, EF are equal and parallel; so also are A C, D F; hence, the two triangles BA C, E D F, having their sides equal, are themselves equal (Prop. XVIII. Bk. I.); consequently, as shown in the last proposition, their planes are parallel. PROPOSITION XVIII.-THEOREM. 427. If two straight lines are cut by three parallel planes, they will be divided proportionally. Let the straight line A B meet the parallel planes, MN, PQ, RS, at the points A, E, B; and the straight line CD meet the same planes at the points C, F, D; then will AE: EB::CF:FD. Draw the line AD, meeting the M A N P E GF R B plane PQ in G, and draw A C, EG, BD. Then the two parallel planes PQ, RS, being cut by the plane ABD, the intersections EG, BD are parallel (Prop. XIII.); and, in the triangle ABD, we have (Prop. XVII. Bk. IV.), AE: EB:: AG: GD. In like manner, the intersections A C, G F being parallel, in the triangle A D C, we have AG: GD::CF:FD; hence, since the ratio A G: G D is common to both pro portions, we have AE: EB::CF: FD. PROPOSITION XIX.-THEOREM. 428. The sum of any two of the plane angles which form a triedral angle is greater than the third. The proposition requires demonstration only when the plane angle, which is compared to the sum of the other two, is greater than either of them. Let the triedral angle whose vertex is S be formed by the three plane angles ASB, ASC, BSC; and suppose the angle ASB to A S D B be greater than either of the other two; then the angle ASB is less than the sum of the angles AS C, BSC. In the plane ASB make the angle BSD equal to BSC; draw the straight line ADB at pleasure; make SC equal S D, and draw A C, B C. The two sides BS, SD are equal to the two sides BS, SC, and the angle BSD is equal to the angle BSC; therefore the triangles BSD, BSC are equal (Prop. V. Bk. I.); hence the side BD is equal to the side B C. But AB is less than the sum of AC and BC; taking BD from the one side, and from the other its equal, BC, there remains AD less than AC. The two sides A S, SD of the triangle ASD, are equal to the two sides A S, SC, of the triangle ASC, and the third side AD is less than the third side AC; hence the angle ASD is less than the angle ASC (Prop. XVII. Bk. I.). Adding BSD to one, and its equal, BS C, to the other, we shall have the sum of ASD, BSD, or AS B, less than the sum of A SC, BSC. |