triangles formed by joining the extremities of these lines; then will these triangles be equal, and their planes parallel.

P

M

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

For, since BE is equal and parallel to AD, the figure ABED is a parallelogram; hence, the side A B 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, DF; 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.

E

R

PROPOSITION XVIII.-THEOREM.

427. If two straight lines are cut by three parallel planes, they will be divided proportionally.

B:

M

P

E

B

AE: EB:: CF: FD.

Draw the line AD, meeting the plane PQ in G, and draw A C, EG, BD. Then the two parallel planes PQ, RS, being cut by the plane A B D, 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 AD C, we have

AG: GD:: CF: FD;

hence, since the ratio A G: G D is common to both proportions, 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.

S

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

be greater than either of the other two; then the angle ASB is less than the sum of the angles A SC, BSC.

A

D

B

In the plane ASB make the angle BSD equal to BSC; draw the straight line ADB at pleasure; make SC equal SD, 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, BS C 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 A D 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, BSC, to the other, we shall have the sum of ASD, BSD, or AS B, less than the sum of A SC, BS C.

PROPOSITION XX.-THEOREM.

429. The sum of the plane angles which form any polyedral angle is less than four right angles.

S

Let the polyedral angles whose vertex is S be formed by any number of plane angles, ASB, BSC, CSD, &c.; the sum of all these plane angles is less than four right angles.

Let the planes forming the poly- A edral angle be cut by any plane,. ABCDEF. From any point, O,

G

D

B

in this plane, draw the straight lines A O, BO, CO, DO, EO, FO. The sum of the angles of the triangles ASB, BSC, &c. formed about the vertex S, is equal to the sum of the angles of an equal number of triangles AOB, BOC, &c. formed about the point O. But at the point B the sum of the angles ABO, OB C, equal to A B C, is less than the sum of the angles ABS, SBC (Prop. XIX.); in the same manner, at the point C we have the sum of BCO, OCD less than the sum of BCS, SCD; and so with all the angles at the points D, E, &c. Hence, the sum of all the angles at the bases of the triangles whose vertex is 0, is less than the sum of all the angles at the bases of the triangles whose vertex is S; therefore, to make up the deficiency, the sum of the angles formed about the point O is greater than the sum of the angles formed about the point S. But the sum of the angles about the point O is equal to four right angles (Prop. IV. Cor. 2, Bk. I.); therefore the sum of the angles about S must be less than four right angles.

430. Scholium. This demonstration supposes that the polyedral angle is convex; that is, that no one of the faces would, on being produced, cut the polyedral angle; if it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude.

PROPOSITION XXL-TEELEM.

4.1. If two triedral angles are formed by plane angles which are equal each to each, the planes of the equal angles will be equally inclined to each other.

Let the two triedral angles whose vertexes are S and T, have the angle ASC equal to DTF, the angle ASB equal to DTE, and the angle BSC equal to ETF; then will the inclination of the planes ASC, ASB be equal to that of the planes DTF, DTE.

P

S

A

B

D

T

^

For, take SB at pleasure; draw BO perpendicular to the plane ASC; from the point 0, at which the perpendicular meets the plane, draw OA, OC, perpendicular to SA, SC; and join A B, BC. Next, take TE equal SB; draw EP perpendicular to the plane DTE; from the point P draw PD, P F, perpendicular respectively to TD, TF; and join DE, EF.

The triangle SA B is right-angled at A, and the triangle TDE at D; and since the angle ASB is equal to DTE, we have SBA equal to TED. Also, SB is equal to TE; therefore the triangle SAB is equal to TDE; hence SA is equal to T D, and A B is equal to DE.

In like manner it may be shown that SC is equal to TF, and BC is equal to EF. We can now show that the quadrilateral ASCO is equal to the quadrilateral DTFP; for, place the angle ASC upon its equal DTF; since SA is equal to TD, and SC is equal to T F, the point A will fall on D, and the point C on F; and, at the same time, AO, which is perpendicular to SA, will fall on DP, which is perpendicular to TD, and, in like manner, CO on FP; wherefore the point O will fall on the point P, and AO will be equal to D P.

But the triangles AOB, DPE are right-angled at O and P; the hypotenuse A B is equal to D E, and the side AO is equal to DP; hence the two triangles are equal (Prop. XIX. Bk. I.); and, consequently, the angle OAB is equal to the angle PDE. The angle OA B is the inclination of the two planes ASB, ASC; and the angle PDE is that of the two planes DTE, DTF; hence, those two inclinations are equal to each other.

432. Scholium 1. It must, however, be observed, that the angle A of the right-angled triangle O A B is properly the inclination of the two planes AS B, ASC only when the perpendicular B O falls on the same side of SA with SC; for if it fell on the other side, the angle of the two planes would be obtuse, and joined to the angle A of the triangle O A B it would make two right angles. But, in the same case, the angle of the two planes DTE, DTF would also be obtuse, and joined to the angle D of the triangle D PE it would make two right angles; and the angle A being thus always equal to the angle D, it would follow in the same manner that the inclination of the two planes ASB, ASC must be equal to that of the two planes D TE, DTF.

433. Scholium 2. If two triedral angles are formed by three plane angles respectively equal to each other, and if at the same time the equal or homologous angles are similarly situated, the two angles are equal. For, by the proposition, the planes which contain the equal angles of the triedral angles are equally inclined to each other.

434. Scholium 3. When the equal plane angles forming the two triedral angles are not similarly situated, these angles are equal in all their constituent parts, but, not admitting of superposition, are said to be equal by symmetry, and are called symmetrical angles.