# 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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