# Introduction, circle geometry

 Page 3 / 3

## Circles iv

1. Find the values of the unknown letters.

Theorem 9 Two tangents drawn to a circle from the same point outside the circle are equal in length.

Proof :

Consider a circle, with centre $O$ . Choose a point $P$ outside the circle. Draw two tangents to the circle from point $P$ , that meet the circle at $A$ and $B$ . Draw lines $OA$ , $OB$ and $OP$ . The aim is to prove that $AP=BP$ . In $▵OAP$ and $▵OBP$ ,

1. $OA=OB$ (radii)
2. $\angle OAP=\angle OPB={90}^{\circ }$ ( $OA\perp AP$ and $OB\perp BP$ )
3. $OP$ is common to both triangles.

$▵OAP\equiv ▵OBP$ (right angle, hypotenuse, side) $\therefore AP=BP$

## Circles v

1. Find the value of the unknown lengths.

Theorem 10 The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment.

Proof :

Consider a circle, with centre $O$ . Draw a chord $AB$ and a tangent $SR$ to the circle at point $B$ . Chord $AB$ subtends angles at points $P$ and $Q$ on the minor and major arcs, respectively. Draw a diameter $BT$ and join $A$ to $T$ . The aim is to prove that $\stackrel{^}{APB}=\stackrel{^}{ABR}$ and $\stackrel{^}{AQB}=\stackrel{^}{ABS}$ . First prove that $\stackrel{^}{AQB}=\stackrel{^}{ABS}$ as this result is needed to prove that $\stackrel{^}{APB}=\stackrel{^}{ABR}$ .

$\begin{array}{ccc}\hfill \stackrel{^}{ABS}+\stackrel{^}{ABT}& =& {90}^{\circ }\left(\mathrm{TB}\perp \mathrm{SR}\right)\hfill \\ \hfill \stackrel{^}{BAT}& =& {90}^{\circ }\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s at centre}\right)\hfill \\ \hfill \therefore \stackrel{^}{ABT}+\stackrel{^}{ATB}& =& {90}^{\circ }\phantom{\rule{3pt}{0ex}}\left(\mathrm{sum of angles in}\phantom{\rule{2pt}{0ex}}▵\mathrm{BAT}\right)\hfill \\ \hfill \therefore \stackrel{^}{ABS}& =& \stackrel{^}{ABT}\hfill \\ \hfill \mathrm{However,}\phantom{\rule{2pt}{0ex}}\stackrel{^}{\mathrm{AQB}}& =& \stackrel{^}{ATB}\phantom{\rule{3pt}{0ex}}\left(\mathrm{angles subtended by same chord}\phantom{\rule{2pt}{0ex}}\mathrm{AB}\right)\hfill \\ \hfill \therefore \stackrel{^}{AQB}& =& \stackrel{^}{ABS}\hfill \\ \hfill \\ \hfill \stackrel{^}{SBQ}+\stackrel{^}{QBR}& =& {180}^{\circ }\phantom{\rule{1.em}{0ex}}\left(\mathrm{SBT}\phantom{\rule{2pt}{0ex}}\mathrm{is a str. line}\right)\hfill \\ \hfill \stackrel{^}{APB}+\stackrel{^}{AQB}& =& {180}^{\circ }\phantom{\rule{1.em}{0ex}}\left(\mathrm{ABPQ}\phantom{\rule{2pt}{0ex}}\mathrm{is a cyclic quad}\right)\hfill \\ \hfill \therefore \stackrel{^}{SBQ}+\stackrel{^}{QBR}& =& \stackrel{^}{APB}+\stackrel{^}{AQB}\hfill \\ \hfill \stackrel{^}{\mathrm{AQB}}& =& \stackrel{^}{ABS}\hfill \\ \hfill \therefore \stackrel{^}{APB}& =& \stackrel{^}{ABR}\hfill \end{array}$

## Circles vi

1. Find the values of the unknown letters.

Theorem 11 (Converse of [link] ) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle.

Proof :

Consider a circle, with centre $O$ and chord $AB$ . Let line $SR$ pass through point $B$ . Chord $AB$ subtends an angle at point $Q$ such that $\stackrel{^}{ABS}=\stackrel{^}{AQB}$ . The aim is to prove that $SBR$ is a tangent to the circle. By contradiction. Assume that $SBR$ is not a tangent to the circle and draw $XBY$ such that $XBY$ is a tangent to the circle.

$\begin{array}{ccc}\hfill \stackrel{^}{ABX}& =& \stackrel{^}{AQB}\phantom{\rule{1.em}{0ex}}\left(\mathrm{tan}-\mathrm{chord theorem}\right)\hfill \\ \hfill \mathrm{However},\phantom{\rule{1.em}{0ex}}\stackrel{^}{\mathrm{ABS}}& =& \stackrel{^}{AQB}\phantom{\rule{1.em}{0ex}}\left(\mathrm{given}\right)\hfill \\ \hfill \therefore \stackrel{^}{ABX}& =& \stackrel{^}{ABS}\hfill \\ \hfill \mathrm{But},\phantom{\rule{1.em}{0ex}}\stackrel{^}{\mathrm{ABX}}& =& \stackrel{^}{ABS}+\stackrel{^}{XBS}\hfill \\ \hfill \mathrm{can only be true if},\phantom{\rule{1.em}{0ex}}\stackrel{^}{\mathrm{XBS}}& =& 0\hfill \end{array}$

If $\stackrel{^}{XBS}$ is zero, then both $XBY$ and $SBR$ coincide and $SBR$ is a tangent to the circle.

1. Show that Theorem [link] also applies to the following two cases:

$BD$ is a tangent to the circle with centre $O$ . $BO\perp AD$ . Prove that:
1. $CFOE$ is a cyclic quadrilateral
2. $FB=BC$
3. $▵COE///▵CBF$
4. $C{D}^{2}=ED.AD$
5. $\frac{OE}{BC}=\frac{CD}{CO}$

1. $\begin{array}{ccc}\hfill \stackrel{^}{FOE}& =& {90}^{\circ }\left(\mathrm{BO}\perp \mathrm{OD}\right)\hfill \\ \hfill \stackrel{^}{FCE}& =& {90}^{\circ }\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{subtended by diameter}\phantom{\rule{2pt}{0ex}}\mathrm{AE}\right)\hfill \\ & \therefore & CFOE\phantom{\rule{2pt}{0ex}}\mathrm{is a cyclic quad}\phantom{\rule{2pt}{0ex}}\left(\mathrm{opposite}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s supplementary}\right)\hfill \end{array}$
2. Let $\stackrel{^}{OEC}=x$ .

$\begin{array}{ccc}& \therefore & \stackrel{^}{FCB}=x\phantom{\rule{3pt}{0ex}}\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{between tangent}\mathrm{BD}\phantom{\rule{2pt}{0ex}}\mathrm{and chord}\phantom{\rule{2pt}{0ex}}\mathrm{CE}\right)\hfill \\ & \therefore & \stackrel{^}{BFC}=x\phantom{\rule{2pt}{0ex}}\left(\mathrm{exterior}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{to cyclic quad}\phantom{\rule{2pt}{0ex}}\mathrm{CFOE}\right)\hfill \\ & \therefore & BF=BC\phantom{\rule{2pt}{0ex}}\left(\mathrm{sides opposite equal}\phantom{\rule{2pt}{0ex}}\angle \mathrm{\text{'}s in isosceles}\phantom{\rule{2pt}{0ex}}▵\mathrm{BFC}\right)\hfill \end{array}$
3. $\begin{array}{ccc}\hfill \stackrel{^}{CBF}& =& {180}^{\circ }-2x\phantom{\rule{3pt}{0ex}}\left(\mathrm{sum of}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{BFC}\right)\hfill \\ \hfill OC& =& OE\phantom{\rule{2pt}{0ex}}\left(\mathrm{radii of circle}\phantom{\rule{2pt}{0ex}}\mathrm{O}\right)\hfill \\ & \therefore & \stackrel{^}{ECO}=x\phantom{\rule{3pt}{0ex}}\left(\mathrm{isosceles}\phantom{\rule{2pt}{0ex}}▵\mathrm{COE}\right)\hfill \\ & \therefore & \stackrel{^}{COE}={180}^{\circ }-2x\phantom{\rule{3pt}{0ex}}\left(\mathrm{sum of}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{COE}\right)\hfill \end{array}$
• $\stackrel{^}{COE}=\stackrel{^}{CBF}$
• $\stackrel{^}{ECO}=\stackrel{^}{FCB}$
• $\stackrel{^}{OEC}=\stackrel{^}{CFB}$
$\begin{array}{ccc}& \therefore & ▵COE|||▵CBF\phantom{\rule{2pt}{0ex}}\left(3\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s equal}\right)\hfill \end{array}$
1. In $▵EDC$

$\begin{array}{ccc}\hfill \stackrel{^}{CED}& =& {180}^{\circ }-x\phantom{\rule{2pt}{0ex}}\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s on a str. line}\mathrm{AD}\right)\hfill \\ \hfill \stackrel{^}{ECD}& =& {90}^{\circ }-x\phantom{\rule{2pt}{0ex}}\left(\mathrm{complementary}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s}\right)\hfill \end{array}$
2. In $▵ADC$

$\begin{array}{ccc}\hfill \stackrel{^}{ACE}& =& {180}^{\circ }-x\left(\mathrm{sum of}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s}\phantom{\rule{2pt}{0ex}}\stackrel{^}{\mathrm{ACE}}\phantom{\rule{2pt}{0ex}}\mathrm{and}\phantom{\rule{2pt}{0ex}}\stackrel{^}{\mathrm{ECO}}\right)\hfill \\ \hfill \stackrel{^}{CAD}& =& {90}^{\circ }-x\phantom{\rule{2pt}{0ex}}\left(\mathrm{sum of}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{CAE}\right)\hfill \end{array}$
3. Lastly, $\stackrel{^}{ADC}=\stackrel{^}{EDC}$ since they are the same $\angle$ .

4. $\begin{array}{ccc}& \therefore & ▵ADC|||▵CDE\phantom{\rule{2pt}{0ex}}\left(3\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s equal}\right)\hfill \\ & \therefore & \frac{ED}{CD}=\frac{CD}{AD}\hfill \\ & \therefore & C{D}^{2}=ED.AD\hfill \end{array}$
1. $\begin{array}{ccc}\hfill OE& =& CD\phantom{\rule{2pt}{0ex}}\left(▵\mathrm{OEC}\phantom{\rule{2pt}{0ex}}\mathrm{is isosceles}\right)\hfill \end{array}$
2. In $▵BCO$

$\begin{array}{ccc}\hfill \stackrel{^}{OCB}& =& {90}^{\circ }\phantom{\rule{2pt}{0ex}}\left(\mathrm{radius}\phantom{\rule{2pt}{0ex}}\mathrm{OC}\phantom{\rule{2pt}{0ex}}\mathrm{on tangent}\phantom{\rule{2pt}{0ex}}\mathrm{BD}\right)\hfill \\ \hfill \stackrel{^}{CBO}& =& {180}^{\circ }-2x\phantom{\rule{2pt}{0ex}}\left(\mathrm{sum of}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{BFC}\right)\hfill \end{array}$
3. In $▵OCD$

$\begin{array}{ccc}\hfill \stackrel{^}{OCD}& =& {90}^{\circ }\phantom{\rule{3pt}{0ex}}\left(\mathrm{radius}\phantom{\rule{2pt}{0ex}}\mathrm{OC}\phantom{\rule{2pt}{0ex}}\mathrm{on tangent}\phantom{\rule{2pt}{0ex}}\mathrm{BD}\right)\hfill \\ \hfill \stackrel{^}{COD}& =& {180}^{\circ }-2x\phantom{\rule{2pt}{0ex}}\left(\mathrm{sum of}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{OCE}\right)\hfill \end{array}$
4. Lastly, $OC$ is a common side to both $▵$ 's.

5. $\begin{array}{ccc}& \therefore & ▵BOC|||▵ODC\phantom{\rule{2pt}{0ex}}\left(\mathrm{common side and 2 equal angles}\right)\hfill \\ & \therefore & \frac{CO}{BC}=\frac{CD}{CO}\hfill \\ & \therefore & \frac{OE}{BC}=\frac{CD}{CO}\left(\mathrm{OE}=\mathrm{CD}\phantom{\rule{2pt}{0ex}}\mathrm{isosceles}\phantom{\rule{2pt}{0ex}}▵\mathrm{OEC}\right)\hfill \end{array}$

$FD$ is drawn parallel to the tangent $CB$ Prove that:
1. $FADE$ is cyclic
2. $▵AFE|||▵CBD$
3. $\frac{FC.AG}{GH}=\frac{DC.FE}{BD}$

1. Let $\angle BCD=x$

$\begin{array}{ccc}& \therefore & \angle CAH=x\phantom{\rule{2pt}{0ex}}\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{between tangent BC and chord CE}\right)\hfill \\ & \therefore & \angle FDC=x\phantom{\rule{2pt}{0ex}}\left(\mathrm{alternate}\angle ,\mathrm{FD}\parallel \mathrm{CB}\right)\hfill \\ & \therefore & \mathrm{FADE}\phantom{\rule{2pt}{0ex}}\mathrm{is a cyclic quad}\phantom{\rule{2pt}{0ex}}\left(\mathrm{chord FE subtends equal}\phantom{\rule{2pt}{0ex}}\angle \mathrm{\text{'}s}\right)\hfill \end{array}$
1. Let $\angle FEA=y$

$\begin{array}{ccc}& \therefore & \angle FDA=y\phantom{\rule{3pt}{0ex}}\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s subtended by same chord}\mathrm{AF}\phantom{\rule{2pt}{0ex}}\mathrm{in cyclic quad}\phantom{\rule{2pt}{0ex}}\mathrm{FADE}\right)\hfill \\ & \therefore & \angle CBD=y\phantom{\rule{2pt}{0ex}}\left(\mathrm{corresponding}\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s,}\phantom{\rule{2pt}{0ex}}\mathrm{FD}\parallel \mathrm{CB}\right)\hfill \\ & \therefore & \angle FEA=\angle CBD\hfill \end{array}$
2. $\begin{array}{ccc}\hfill \angle BCD& =& \angle FAE\phantom{\rule{2pt}{0ex}}\left(\mathrm{above}\right)\hfill \end{array}$
3. $\begin{array}{ccc}\hfill \angle AFE& =& {180}^{\circ }-x-y\phantom{\rule{2pt}{0ex}}\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{AFE}\right)\hfill \\ \hfill \angle CBD& =& {180}^{\circ }-x-y\phantom{\rule{2pt}{0ex}}\left(\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s in}\phantom{\rule{2pt}{0ex}}▵\mathrm{CBD}\right)\hfill \\ & \therefore & ▵AFE|||\phantom{\rule{2pt}{0ex}}▵CBD\phantom{\rule{2pt}{0ex}}\left(3\phantom{\rule{2pt}{0ex}}\angle \phantom{\rule{2pt}{0ex}}\mathrm{\text{'}s equal}\right)\hfill \end{array}$
1. $\begin{array}{ccc}\hfill \frac{DC}{BD}& =& \frac{FA}{FE}\hfill \\ & \therefore & \frac{DC.FE}{BD}=FA\hfill \end{array}$
2. $\begin{array}{ccc}\hfill \frac{AG}{GH}& =& \frac{FA}{FC}\phantom{\rule{2pt}{0ex}}\left(\mathrm{FG}\parallel \mathrm{CH}\phantom{\rule{2pt}{0ex}}\mathrm{splits up lines}\phantom{\rule{2pt}{0ex}}\mathrm{AH}\phantom{\rule{2pt}{0ex}}\mathrm{and}\phantom{\rule{2pt}{0ex}}\mathrm{AC}\phantom{\rule{2pt}{0ex}}\mathrm{proportionally}\right)\hfill \\ & \therefore & FA=\frac{FC.AG}{GH}\hfill \end{array}$
3. $\begin{array}{ccc}& \therefore & \frac{FC.AG}{GH}=\frac{DC.FE}{BD}\hfill \end{array}$

How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
Got questions? Join the online conversation and get instant answers!  By By    By  By Abishek Devaraj By