PROPOSITION XVI. THEOREM.

If a line is perpendicular to a plane, every plane passed through the perpendicular, is also perpendicular to the plane.

A

Let AP be perpendicular to the plane NM; then will every plane passing through AP be perpendicular to NM. Let BF be any plane passing through AP, and BC its intersection with the plane MN. In the plane MN, draw DE perpendicular to BP: then the line AP, M being perpendicular to the plane MN, is perpendicular to each of the two straight lines BC, DE. Now, since AP and DE are both perpendicular to the common intersection BC, the angle which they form will measure the angle between the planes (D. 4): but the angle APD, or APE, is a right angle: hence, the two planes are perpendicular to each other.

i

B

E

P

N

D

Scholium. When three straight lines, such as AP, BP, DP, are perpendicular to each other, any two may be regarded as determining a plane, and the three will determine three planes. Now, each line is perpendicular to the plane of the other two, and the three planes are perpendicular to each other.

་

PROPOSITION XVII. THEOREM.

Conversely: If two planes are perpendicular to each other, a line drawn in one of them perpendicular to their common intersection, will be perpendicular to the other plane.

Let the plane BA be perpendicular to NM; then, if the line AP be perpendicular to the intersection BC, it will also be perpendicular to the plane NM.

For, in the plane MN, draw PD perpendicular to PB; then, because the planes are perpendicular, the angle APD

is a right angle (D. 4); therefore, the line

M

A

B

P

D

N

AP is perpendicular to the two straight lines PB, PD, passing through its foot; therefore, it is perpendicular to their plane MN (p. 4).

Cor. If the plane BA is perpendicular to the plane MN, and if at a point P of the common intersection we erect a perpendicular to the plane MN, that perpendicular will be in the plane BA. For, let us suppose it will not, then, in the plane BA draw AP perpendicular to PB, the common intersection, and this AP at the same time, is perpendicular to the plane MN, by the theorem; therefore at the same point P there are two perpendiculars to the plane MN, one out of the plane BA, and one in it, which is impossible (P. 4, c. 2).

PROPOSITION XVIII. THEOREM.

If two planes which cut each other are perpendicular to a third plane, their common intersection is also perpendicular to that plane.

Let the planes BA, DA, be perpendicular to NM; then will their intersection AP be perpendicular to NM.

For, at the point P, erect a perpendicular to the plane MN; that perpendicular must be at once in the plane AB and in the plane AD (P, 17, c.); therefore, it is their common intersection AP

M

A

F

B

E

P

N

PROPOSITION XIX. THEOREM.

The sum of either two of the plane angles which include a triedral angle, is less than the third.

The proposition requires demonstration only when the plane angle, which is compared with the sum of the two others, is greater than either of them. Therefore, suppose the triedral angle S to be formed by the three plane angles ASB, ASC, BSC, and that the angle ASB is the greatest; we are to show that ASB<ASC+BSC.

In the plane ASB make the angle_BSD=BSC, and draw the straight line ADB at pleasure; then make SC SD, and draw AC, BC.

A

S

D

B

hence, BD-BC. But

The two sides BS, SD, are equal to the two BS, SC, and the angle BSD=BSC; therefore, the triangles BSD, BSC, are equal (B. I., P. 5); AB<AC+BC; taking BD from the one side, and from the other its equal BC, there remains AD AC. The two sides AS, SD, are equal to the two AS, SC; the third side AD is less than the third side AC; therefore, the angle ASD< ASC (B. I., P. 9, c.) Adding BSD=BSC, we have

ASD+BSD, or ASB <ASC+BSC.

PROPOSITION XX. THEOREM.

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

Let S be the vertex of a polyedral angle bounded by the faces BSC, CSD, DSE, ESA, ASB; then will the sum of the plane angles about S be less than four right angles. For, let the polyedral angle be cut

by any plane AD, intersecting the edges in the points A, B, C, D, E, and the faces in the lines AB, BC, CD, DE, EA. From any point of this plane, as O, draw the straight lines OA, OB, OC, OD, OE.

E

A

B

S

We thus form two sets of triangles, one set having a common vertex S, the other having a common vertex 0, and both having the common bases AB, BC, CD, DE, EA. Now, in the set which has the common vertex S, the sum of all the angles is equal to the sum of all the plane angles which comprise the polyedral angle whose vertex is S, together with the sum of all the angles at the bases: viz.: SAB, SBA, SBC, &c.; and the entire sum is equal to twice as many right angles as there are triangles. In the set whose

common vertex is 0, the sum of all the angles is equal to the four right angles about O, together with the interior angles of the polygon, and this sum is equal to twice as many right angles as there are triangles. Since the number of triangles, in each set, is the same, it follows that these sums are equal. But in the triedral angle whose vertex is B, ABS+SBQ>ABC (P. 19), and the same may be shown at each of the other vertices, C, D, E, A: hence, the sum of the angles at the bases, in the triangles whose common vertex is S, is greater than the sum of the angles at the bases, in the set whose common vertex is 0: therefore, the sum of the vertical angles about S is less than the sum of the angles about 0: that is, less than four right angles.

Scholium. This demonstration is founded on the supposition that the polyedral angle is convex, or that the plane of no one face produced can ever meet 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 XXI. THEOREM.

If two triedral angles are included by plane angles which are equal each to each, the planes of the angles are equally inclined to each other.

Let S and T be the vertices of two triedral angles, and let the angle ASC=DTF, the angle ASB=DTE, and the angle BSC-ETF; then will the inclination of the planes ASC, ASB, be equal to that of the planes DTF, DTE. For, having taken SB at

pleasure, draw BO perpendicu

S

T

TE=SB; draw EP perpendicular to the plane DTF; from the point P draw PD, PF, perpendicular respectively to TD, TF; lastly, draw DE and EF.

In like manner, it may be shown, that SC=TF, and BC=EF. That proved, the quadrilateral ASCO is equal to the quadrilateral DTFP: for, place the angle ASC upon its equal DTF; because SA=TD, and_SC=TF, 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, OC on PF; wherefore, the point O will fall on the point P, and hence, AO is equal to DP.

But the triangles AOB, DPE, are right-angled at O and P; the hypothenuse AB-DE, and the side A0=DP: hence, those triangles are equal (B. I., P. 17); and, consequently, the angle OAB=PDE. But the angle OAB measures the inclination of the two faces ASB, ASC; and the angle PDE measures that of the two faces DTE, DTF; hence, those two inclinations are equal to each other.

Scholium 1. It must, however, be observed, that the angle A of the right-angled triangle AOB is properly the inclination of the two planes ASB, ASC, only when the perpendicular BO 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 the obtuse angle together with the angle A of the triangle OAB would make two right angles. But in the same case, the angle of the two planes TDE, TDF, would also be obtuse, and the obtuse angle together with the angle D of the triangle DPE, would make two right angles; and the angle A being thus always equal to the angle D, it would follow that the inclination of the two planes ASB, ASC, must be equal to that of the two planes TDE, TDF.

Scholium 2. If two triedral angles are included by three