|
|
|
Viewing Version
7
of
'circumferential angle is half the corresponding central angle'
|
[ view 'circumferential angle is half the corresponding central angle'
|
back to history
]
| Title of object: |
circumferential angle is half the corresponding central angle |
| Canonical Name: |
CircumferentialAngleIsHalfCorrespondingCentralAngle |
| Type: |
Theorem |
| Created on: |
2007-06-07 18:14:48 |
| Modified on: |
2007-06-07 19:04:47 |
| Classification: |
msc:51M04 |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
\usepackage{pstricks}
% there are many more packages, add them here as you need them
% define commands here
\newtheorem{thm}{Theorem}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{ax}{Axiom} |
Content:
\begin{thm} \emph{[Euclid, Book III, Prop. 20]} In any circle, a circumferential angle is half the size of the central angle subtending the same arc.
\end{thm}
\begin{proof}
There are actually several distinct cases. Consider $\angle BAC$ in a circle with center $O$, and draw $AO, BO, CO$ as well as the chord containing both $A$ and $O$:
\begin{center}
% Generated by eukleides 1.0.3
%frame(-1,-2,5,3.2)
%A B C triangle(4.2)
%c=circle(A,B,C)
%draw(c)
%draw(A,B,C)
%draw("A",A,-180:)
%draw("B",B,0:)
%draw("C",C,90:)
%O=center(c)
%draw("O",O,140:)
%draw(O,dot)
%F' F intersection(line(A,O),c)
%draw(segment(F',F))
%draw("F",F,0:)
%a=segment(O,A)
%b=segment(O,B)
%c=segment(O,C)
%draw(b)
%draw(c)
%mark(a)
%mark(b)
%mark(c)
\begin{pspicture*}(-1.0000,-2.0000)(5.0000,3.2000)
\rput(-1.1,-2){.}
\rput(5,4){.}
\pscircle(2.1000,0.4822){2.1547}
\pspolygon(0.0000,0.0000)(4.2000,0.0000)(1.5750,2.5720)
\uput{0.3000}[-180.0000](0.0000,0.0000){A}
\uput{0.3000}[0.0000](4.2000,0.0000){B}
\uput{0.3000}[90.0000](1.5750,2.5720){\PMlinkescapetext{C}}
\uput{0.3000}[140.0000](2.1000,0.4822){O}
\psdots[dotstyle=*, dotscale=1.0000](2.1000,0.4822)
\psline(0.0000,0.0000)(4.2000,0.9645)
\uput{0.3000}[0.0000](4.2000,0.9645){F}
\psline(2.1000,0.4822)(4.2000,0.0000)
\psline(2.1000,0.4822)(1.5750,2.5720)
\psline(1.0836,0.0949)(1.0164,0.3873)
\psline(3.1836,0.3873)(3.1164,0.0949)
\psline(1.6920,1.4906)(1.9830,1.5637)
\end{pspicture*}
% End of figure
\end{center}
In this case, the center of the circle lies between the arms of the circumferential angle. Now, since $AO=OB$, $\triangle AOB$ is isosceles, and $\angle FOB$ is an exterior angle. Thus
\[\angle FOB=\angle OAB + \angle OBA = 2\angle OAB\]
Similarly, $\triangle AOC$ is isosceles, and
\[\angle FOC=\angle OAC + \angle OCA = 2\angle OAC\]
and it follows that
\[\angle BOC=\angle FOB + \angle FOC = 2\angle OAB + 2\angle OAC = 2\angle BAC\]
proving the result.
A second case is the case in which both arms of the angle lie to one side of the circle's center:
\begin{center}
% Generated by eukleides 1.0.3
%frame(-1,-3,5,2)
%A B C triangle(4.2,30:,40:)
%c=circle(A,B,C)
%draw(c)
%draw(A,B,C)
%draw("A",A,-180:)
%draw("B",B,0:)
%draw("C",C,90:)
%O=center(c)
%draw("O",O,-45:)
%draw(O,dot)
%F' F intersection(line(A,O),c)
%draw(segment(F',F))
%draw("F",F,0:)
%a=segment(O,A)
%b=segment(O,B)
%c=segment(O,C)
%draw(b)
%draw(c)
%mark(a)
%mark(b)
%mark(c)
\begin{pspicture*}(-1.0000,-3.0000)(5.0000,2.0000)
\rput(-1.1,-3){.}
\rput(5,2.1){.}
\pscircle(2.1000,-0.7643){2.2348}
\pspolygon(0.0000,0.0000)(4.2000,0.0000)(2.4881,1.4365)
\uput{0.3000}[-180.0000](0.0000,0.0000){A}
\uput{0.3000}[0.0000](4.2000,0.0000){B}
\uput{0.3000}[90.0000](2.4881,1.4365){\PMlinkescapetext{C}}
\uput{0.3000}[-45.0000](2.1000,-0.7643){O}
\psdots[dotstyle=*, dotscale=1.0000](2.1000,-0.7643)
\psline(0.0000,0.0000)(4.2000,-1.5287)
\uput{0.3000}[0.0000](4.2000,-1.5287){F}
\psline(2.1000,-0.7643)(4.2000,0.0000)
\psline(2.1000,-0.7643)(2.4881,1.4365)
\psline(0.9987,-0.5231)(1.1013,-0.2412)
\psline(3.0987,-0.2412)(3.2013,-0.5231)
\psline(2.1463,0.3621)(2.4418,0.3100)
\end{pspicture*}
% End of figure
\end{center}
The proof is similar to the previous case, except that the angle in question is the difference rather than the sum of two known angles. Here we see that both $\triangle AOB$ and $\triangle AOC$ are isosceles, so that again
\begin{align*}
\angle COF &= 2\angle OAC \\
\angle BOF &= 2\angle OAB
\end{align*}
Subtracting, we get
\[\angle COB = \angle COF - \angle BOF = 2\angle OAC -2\angle OAB = 2\angle BAC\]
as desired.
The final case is the case in which one arm of the angle goes through the center of the circle. This is a degenerate form of the first case, and the same proof follows through except that one of the angles is zero.
\end{proof} |
|
|
|
|
|
