# Introduction, circle geometry

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## 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. $\begin{array}{ccc}& \therefore & \frac{FC.AG}{GH}=\frac{DC.FE}{BD}\hfill \end{array}$

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Joseph
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Lohitha
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nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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da
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Bhagvanji
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Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
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narayan
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ya I also want to know the raman spectra
Bhagvanji
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what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
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Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
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Alexandre
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Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
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What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
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Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
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Rafiq
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Anam
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Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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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
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