For as ADB C, therefore A D B D::B C B D. But A D : BD::AB; and BC BD::C:D (Theo. 73.); hence A: B:: C: D.

Cor. If there be three lines, and the rectangle of the two extremes be equal to the square of the mean, these three lines are proportionals.

THEOREM LXXVII.

If three straight lines, A, B, and C, be proportionals, that is, if A: B :: B: C, then A : C :: A2 : B2.

For A AB: A: B, and B2 Hence, if A B::B: C, then A2

BC: B: C (Theo. 73. Cor.)

A B:: B2 BC;

A

THEOREM LXXVIII.

A line, as BE, drawn parallel to CD, one of the sides of a triangle ACD, divides the other sides A C, AD proportionally, or so that AB: AC::AE: A D.

Let CE and BD be joined; then the triangles,

A

B

E

B CE, B D E, being on the same base B E, and between the same parallels, are equal (Cor. 1. Theo. 33.); and if to each be added the triangle A B E, the triangles A CE and A B D will also be equal. Hence the triangle A B D will have the same proportion to the whole triangle ACD that the triangle A C E has to the whole triangle A CD. But the triangle ABD is to the triangle A C D as A B is to AC; and the triangle ACE is to the triangle A D C as A E is to A D. (Theo. 73.) Hence A B is to A C as A E is to A D.

Cor. A BBC::AE: E D, and B C A CE DA D.

THEOREM LXXIX.

Equiangular triangles, as A B C, DEF have their corresponding sides about the equal angles, as C and F, proportional, viz. CA: CB::FD: FE.

Conceive the point F to be laid on the point C, and the line F D on the line C A, and let the point D fall at G; then as F D coincides with CG, and the angle C is equal to the angle F, the line F E will fall on the line C B. Let the

G

A

D
B

point E fall at H. Now, as the angle F D E, or the angle CG H, is equal to the angle A, the line G H is parallel to the line A B; and hence CA: CB::CG: C H (Theo. 78.), or ::FD: FE.

In the same way it may be shewn that A C; A B::FD: DE, and that A B B C::DE E F.

Cor. Equiangular triangles are similar.

E

If ABC be a triangle, right-angled at C, and CD be a perpendicular from the right angle C, on the hypothenuse A B, then AD: DC:: DC: DB, AB. AC :: AC: AD, and AB . BC:: BC: BD.

DB

For the triangles A B C, A D C, having the common angle A, and the right angles A D C and AC B equal, are similar (Cor. Theo. 79.); and in like manner may BDC be shewn to be similar to AC B. Hence the triangles A B C, A D C, and B D C, are all similar, and consequently A D C D::CD: D B, and A BAC::AC: A D, and A B: BC: BC: CD (Theo. 79.) Cor. 1. AB. A D

= D C2.

= A C2, AB. BD = B C2, and A D.DB

Cor. 2. AB. AD + AB.BD = A C2 + B C2; or, A B2 = A C2+ B C2.

THEOREM LXXXI.

If the angle C, of the triangle ABC, be bisected by the line CD, then will A C be to CB as AD to DB.

But as

E

Let BC, parallel to CD, meet AC produced in E. Then A C CE::AD: D B (Theo. 78.) the parallel lines DC and B E are cut by the line ACE, the angles E and ACD are equal (Theo. 19.); and as the same parallel lines are cut by the line B C, the angles D C B and C B E are equal (Theo. 17.) Hence, as A CD and D C B are equal, E and CBE are equal, and consequently (Theo. 4.) CE is equal to C B; therefore A C : CB::AD: D B.

A DB

THEOREM LXXXII.

F

BD

E

If ABC, DFE be two similar triangles, AB and D E sides opposite the equal angles C and F; the triangle A B C will be to the triangle DFE as the square of AB to the square of D E, or as A B2 to DE2. For let A K, D M, be the squares on A B and DE; BI and E L diagonals of these squares; and C G, F H, perpendiculars from A C and F upon A B and D E. Then, as the angle CA G is equal to the angle F D H, and the angles A C G, D H F, being right angles, are equal to each other, the angles A C G, D F H, are also equal (Theo. 24. Cor 1.); and consequently the triangles AC G and DFH are equiangular. Hence AC: DF:: CG: FH (Theo.79.) But AC : DF:: A B : DE, or :: AI: DL; therefore C G F H :: AI: DL; or CG: AI:: FH: DL. But the triangle A B C is to the

:

Ꮮ M

triangle A BI as C G is to A I; and the triangle D E F is to the triangle D L E as F H is to D L (Theo. 74.); therefore the triangle A B C is to the triangle A B I as the triangle D F E to the triangle DLE; or the triangle A B C is to the triangle D F E as twice the triangle A BI is to twice the triangle D L E, or as A K is to DM; that is, as the square of A B is to the square of D E.

THEOREM LXXXIII.

If ABCDE be an equilateral polygon inscribed in the circle whose centre is M, and FGHIK an equilateral polygon of a like number of sides inscribed in the circle whose centre is L, the perimeter of the polygon ABCDE will be to the perimeter of the polygon FGHIK, as the radius AM to the radius F L.

B

F

G

K

E

M

D

H

For join M to A, B, C, D, and E, and L to F, G, H, I, and K, then the triangles A M B, B M C, &c. being mutually identical; and the triangles F LG, GL H, &c. being also mutually identical (Theo. 5.), the angles A M B, FL G, will be the like parts of four right angles, and they will consequently be equal. Therefore, as the triangles are isoceles, the angles M A B, M B A, will be respectively equal to the angles LG F, LFG. Whence the triangles are equiangular; and therefore M A: FL:: A B : F G (Theo. 79.), or :: the perimeter A B C DEA: the perimeter F G H I K F (Theo. 70.), since the perimeters are like multiples of A B and F G.

I

Cor. If we conceive the sides of the polygons to be indefinitely small, and their number to be indefinitely great, the property which has here been proved of equilateral polygons inscribed in a circle, will become a property of the circles themselves; for the polygons will then coincide with the circumferences of the circles. Hence the circumferences of circles are to each other as their radii.

THEOREM LXXXIV.

If ABC be a triangle, one of whose angles AC B is bisected by the line CE; the rectangle AC, CB is equal to the rectangle AE, E B, together with the square of E C.

Let A CBD, be a circle circumscribing the triangle, and let C E, produced, meet the circumference in D, and join D B. Then the angles CAE and C D B are equal, as they are in the same segment (Cor. 2. Theo. 53.); and the angles A CE and D C B are equal also, because the line CED bisects the angle A C B. Hence the angle

C

A

B

E

D

AEC is equal to the angle D B C (Theo. 24. Cor. 1.), and the triangles AEC and DBC are similar. Therefore AC:CE:: DC: CB (Theo. 79.); and consequently AC.CBEC. CD (Theo. 75.) But E C. C D E C2 + E C. E D (Theo. 40.), and E C.ED = AE.EB (Cor. 1. Theo. 67.) Hence A C. CB = AE.EB + E C2.

THEOREM LXXXV.

=

IfCD be a diameter of a circle, circumscribing the triangle ABC, and CE a perpendicular from the angle C on the opposite side A B; the rectangle A C. CB is equal to the rectangle DC. CE.

A

B

Join D B. Then the angles CA E and C D B are equal, as they are in the same segment (Cor. 2. Theo. 53.), and the angle C B D, in a semicircle, is equal to the right angle C E A (Cor. 3. Theo. 53.); therefore the angles A CE and D C B are equal (Theo. 24. Cor. 1.), and the triangles A EC and D C B are similar. Hence AC:CE::DC: C B (Theo. 79.) ; and consequently A C. C B CD. CE (Theo. 75.)

THEOREM LXXXVI.

A

D

D

E

B

If ABCD be a quadrilateral inscribed in a circle, the rectangle A C. BD is equal to the two rectangles AD. BC, and A B. DC. From C let CE be drawn, making the angle DCE equal to the angle A C B. Then the angles CDE, CA B, standing on the same segments, are equal (Cor. 2. Theo. 53.); and therefore the angles DEC and A B C are equal (Theo. 24. Cor. 1.) Hence the triangles DEC and ABC are similar, and consequently A B : AC :: DE: DC (Theo. 79.) Whence A B . D C= AC.DE (Theo. 75.) Now, if from the equal angles DC E and A CB, the common angle A CE be taken away, the remaining angles DCA and ECB will be equal; and the angles EBC and DAC standing on the same segment are equal, (Cor. 2. Theo. 53.); therefore the angles ADC and BEC are equal, (Theo. 24. Cor. 1.) and, consequently, AD: AC::BE: BC, (Theo. 79.) or AD. BC= AC.BE. (Theo. 75.) But it has been shewn, that AB.DC= A C.DE, therefore A B.DC+AD.BC=AC.DE+AC.BE, or= AC.DB.

THEOREM LXXXVII.

Let A B, an arc of a circle, be bisected in C; draw the chords AB, A C, and BC; and to any point, D in the circumference, draw AD, BD, and CD; then AB: AC::AD+BD: DC.

For (Theo. 86.) AD. CB+BD. AC=AB.CD. But (Theo. 39.) AD.CBBD.CB=AD + BD. CB, or =AD+BD. A C. Hence A B. CD=AD+BD . AC, and therefore (Theo. 76.) A B AC::AD + BD: DC.

If BDFE be a rectangle inscribed in the right-angled triangle ABC, the right angle B being common to both, the rectangle AF. F C is equal to the two rectangles AD. D B and B E. F C together.

Let E G be perpendicular to F C; then the triangles ADF and EFC are similar, and consequently A D

AF

B

:: FG: EF, or B D. Hence A D.BD=AF
.F G. In the same manner A F: FD, or EB
::EC: C G. Hence B E. EC=AF. G C ; A
and therefore AD. BD+BE.EC=AF.FG
+AF. GC, or AF. F C.

=

ON THE INTERSECTION OF PLANES.

DEFINITIONS.

1. A STRAIGHT line is perpendicular to a plane when it makes right angles with every straight line which it meets in that plane.

2. Two planes are perpendicular to each other when any straight line drawn in one of the planes, perpendicular to their common section, is perpendicular to the other plane.

3. If two planes cut each other, and from any point in the line of their common section, two straight lines be drawn, at right angles to that line, one in the one plane, and the other in the other plane, the angle contained by these two lines is the angle made by the planes.

4. A straight line is parallel to a plane when it does not meet the plane though produced ever so far.

5. Planes are parallel to each other when they do not meet, though produced ever so far.

6. A solid angle is one which is formed by the meeting, in one point, of more than two plane angles, which are not in the same plane with each other.

THEOREM LXXXIX.

If any three straight lines, as A B, C D, C B meet one another, as in C, B and E, they are in one plane.