« ForrigeFortsett »
For as AD = B C, therefore A D:BD::BC:B D. But A D :BD::A:B; 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 : Bo.
For A : AB :: A:B, and BP: BC::B:C (Theo. 73. Cor.) Hence, if A:B::B:C, then A : AB::B:BC; AA': Bo :: AB: BC. (Theo. 71.) But (Theo. 73. BCor.) AB:BC::A:C; therefore A:C:: Ao : Bo C
THEOREM LXXVIII. A line, as BE, drawn parallel to CD, one of the sides of a triangle AC D, divides the other sides A C, A D proportionally, or so that A B : AC::AE: A D. Let CE and B D be joined; then the triangles,
А BC E, B D E, being on the same base B E, and between the same parallels, are equal (Cor. 1. Theo. 33.);
В, E and if to each be added the triangle A BE, the triangles A CE and A B D will also be equal. Hence
D the triangle A B D will have the same proportion to the whole triangle ACD that the triangle A CE has to the whole triangle A CD. But the triangle ABD is to the triangle A C D as A B is to A C; and the triangle ACE is to the triangle ADC 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 B:BC::AE: E D, and BC: AC::ED: A D.
THEOREM LXXIX. Equiangular triangles, as ABC, 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
G H CG, and the angle C is equal to the angle F,
DE the line F E will fall on the line C B. Let the
A B point E fall at H. Now, as the angle F D E, or the angle C G H, is equal to the angle A, the line G H is parallel to the line A B; and hence CA:CB::CG:CH (Theo. 78.), or ::FD:FE.
In the same way it may be shewn that AC: AB::FD:D E, and that A B:BC::DE: E F.
Cor. Equiangular triangles are similar.
If A B C be a triangle, right-angled at C, and C D be a perpendicular from the right angle C, on the hypothenuse A B, then A D :DC::DC: D B, AB : AC :: AC: AD, and AB. BC::BC: BD.
For the triangles A BC, ADC, having the common angle A, and the right angles ADC and ACB equal, are similar (Cor. Theo. 179.); and in like
D B Hence the triangles ABC, ADC, and BDC, are all similar, and consequently AD:CD::CD:D B, and AB: AC::AC: A D, and A B:BC::BC:CD (Theo. 79.)
Cor. 1. A B. AD= A C, A B.B D = B C , and AD.DB = D C2.
Cor. 2. A B. AD + A B. BD = A C2 + B C?; or, A B? = AC + BC%.
If the angle C, of the triangle ABC, be bisected by the line CD, then will AC be to CB as A D to D B. Let B C, parallel to CD, meet AC produced in E.
E Then A C:CE :: AD:D B (Theo. 78.)
But as the parallel lines D C 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
A D B 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 C B E are equal, and consequently (Theo. 4.) CE is equal to C B; therefore AC: CB:: AD: D B.
If ABC, DFE be two similar triangles, A B 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 A B to the square of D E, or as A Bo to D E . For let A K, DM, be the squares on A B
F and DE; BI and E L diagonals of these squares; and C G, FH, perpendiculars from A BDA E C and F upon A B and D E. Then, as the angle CAG is equal to the angle FDH, and
L M the angles ACG, DHF, being right angles,
K are equal to each other, the angles ACG, DFH, are also equal (Theo. 24. Cor 1.); and consequently the triangles A CG and DFH are equiangular. Hence AC: DF::CG:FH (Theo.79.) But AC : DF:: AB:DE, or :: AI:DL; therefore CG:FH:: AI: triangle A B I as C G is to Al; and the triangle DEF 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 B I is to twice the triangle DLE, or as A K is to D M; that is, as the square of A B is to the square
of D E.
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 A B C D E will be to the perimeter of the polygon FGHIK, as the radius AM to the radius FL.
For join M to A, B, C, D, and E, and L to F, G, H, I, and K, then the triangles A MB, BMC, &c.
K K mutually identical (Theo. 5.), the angles A MB, FL G, will be the like
H parts of four right angles, and they
I 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 LGF, LFG. Whence the triangles are equiangular; and therefore M A :FL :: A B : F G (Theo. 79.), or :: the perimeter ABC DE A : the perimeter F G HIKF (Theo. 70.), since the perimeters are like multiples of A B and F G.
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.
If A B C be a triangle, one of whose angles AC B is bisected by the line CE; the rectangle AC, CB is equal to the rectangle A E, E B, together with the square of EC.
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
B CA E and C D B are equal, as they are in the
E same segment (Cor. 2. Theo.53.); and the angles AC E and D C B are equal also, because the line
D. CE D bisects the angle A CB. Hence the angle
AEC is equal to the angle D BC (Theo. 24. Cor. 1.), and the triangles A E C and DB C are similar. Therefore A C:CE::DC : CB (Theo. 79.); and consequently A C.C B = EC.CD (Theo. 75.) But E C.CD= E C + E C. E D (Theo. 40.), and E C.E D = AE.EB (Cor. 1. Theo. 67.) Hence AC.CB = AE.EB + E C.
If CD be a diameter of a circle, circumscribing the triangle A B C, 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.
Join D B. Then the angles C A E and C D B are equal, as they are in the same segment (Cor. 2. Theo. 53.), and the angle C BD, in a semicircle, is equal to the right angle CE A (Cor. 3. Theo. А
B 53.); therefore the angles A CE and D C B are equal (Theo. 24. Cor. 1.), and the triangles AEC D and D C B are similar. Hence AC:CE:: DC: C B (Theo.79.); and consequently AC.CB=CD. CE (Theo. 75.)
If ABC D be a quadrilateral inscribed in a circle, the rectangle AC. BD is equal to the two rectangles AD.BC, and AB.DC.
From Clet CE be drawn, making the angle D
B В the triangles DEC and A B C are similar, and consequently A B: AC::DE: DC (Theo. 79.) Whence A B.DC= AC.DE (Theo. 75.) Now, if from the equal angles DCE and ACB, 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 BE C are equal, (Theo. 24. Cor. 1.) and, consequently, AD:AC :: BE : BC, (Theo. 79.) or AD.BC= A C.BE. (Theo. 75.) But it has been shewn, that AB.DC= AC.DE, therefore A B.DC+AD.BC=AC.DE+AC.BE, or= AC.DB.
Let A B, an arc of a circle, be bisected in C; draw the chords AB, AC, and BC; and to any point, D in the circumference, draw AD, For (Theo. 86.) AD.CB+BD.AC=AB.CD. But (Theo. 39.) AD.CB + BD.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 B D F E be a rectangle inscribed in the right-angled triangle ABC, the right angle B being common to both, the rectangle AF.FC is equal to the two rectangles AD.D B and B E F C together.
Let E G be perpendicular to FC; then the triangles ADF and E F C are similar, and consequently A D : AF
B ::FG:E F, or B D. Hence AD.BD= AF
E .FG. In the same manner AF:FD, or E B D ::EC:CG. Hence B E. E C = AF.GC;
A F GC and therefore AD.BD+BE. EC = AF.FG + A F, GC, or = AF.FC.
ON THE INTERSECTION OF PLANES.
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.
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.