A central angle is an angle substended by an arc (or chord) of a circle at the centre.
စက်ဝိုင်းတစ်ခု၏အဝန်းပိုင်းတစ်ခု(သို့မဟုတ်လေးကြိုးတစ်ခု)ကယင်းစက်ဝိုင်း၏ဗဟို၌ခံဆောင်ထားသောထောင့်ကိုဗဟိုခံထောင့်ဟုခေါ်သည်။
An inscribed angle is an angle substended by an arc (or chord) of a circle at a point on the other arc.
စက်ဝိုင်းတစ်ခု၏အဝန်းပိုင်းတစ်ခု(သို့မဟုတ်လေးကြိုးတစ်ခု)ကကျန်အဝန်းပိုင်းပေါ်ရှိအမှတ်တစ်ခု၌ခံဆောင်ထားသောထောင့်ကိုအဝန်းခံထောင့်ဟုခေါ်သည်။
∠ COD is a central angle substended by arc CD (or chord CD).
∠ PQR is an inscribed angle substended by arc PR (or chord PR).
Cyclic Quadrilateral
A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral.
စတုဂံတစ်ခု၏ထောင့်စွန်းလေးခုစလုံးသည်စက်ဝိုင်းတစ်ခုတည်းပေါ်ကျရောက်နေသောစတုဂံကိုစက်ဝိုင်းတွင်းကျစတုဂံဟုခေါ်၏။
9.2 Properties of Chords
Problems
Example 1
In the given figure, O is the centre of the circle, arc PQ= arc QR= arc RS. Find PR.
$\begin{aligned}
\operatorname{arcPQ}&=\operatorname{arc} Q R=\operatorname{arc} R S\quad(\text { given }) \\
\therefore \angle P O Q&=\angle Q O R=\angle R O S \quad(\text { Theorem } 3) \\
\text { But } \angle P O Q&+\angle Q O R+\angle R O S=90^{\circ} \\
\therefore \angle P O Q&=\angle Q O R=\angle R O S=\frac{90^{\circ}}{3}=30^{\circ} \\
\angle P O R&=\angle P O Q+\angle Q O R=30^{\circ}+30^{\circ}=60^{\circ} \\
\angle P&=\angle R \quad(\because O R=O P \quad\text { radii }) \\
\therefore \angle P&=\angle R=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ} \\
\therefore \triangle &P O R \text { is equalateral. } \\
\therefore P R&=O R \\
\text { But } O R&=O S \quad(\because \text { radii }) \\
\therefore P R&=O S=3 \mathrm{~cm}
\end{aligned}$
Example 2
In ⊙ O, find the values of α, β and θ .
(a)
$\begin{aligned}
\angle O A B&=\angle O B A\quad(\because O A= O B \text { radii }) \\
\therefore \angle O A B&=40^{\circ} \\
\alpha&=180^{\circ}-\left(40^{\circ}+40^{\circ}\right)=100^{\circ} \\
\alpha&=2 \beta\quad(\because \text { angles substended by arc } A B) \\
\beta&=\frac{1}{2} \alpha=\frac{1}{2}\left(100^{\circ}\right)=50^{\circ} \\
\theta&=\beta\quad(\because \text { angles substended by } \operatorname{arc} A B) \\
\therefore \theta&=50^{\circ} \\
\end{aligned}$
(b)
$\begin{aligned}
\beta&=\dfrac{1}{2} \times 200^{\circ}\quad(\because \text { angles substended by arc } A C B) \\
\beta&=100^{\circ} \\
\alpha+\beta&=180^{\circ}\quad(\because \text { opposite angles of cyclic quadrilateral } A D B C) \\
\alpha&=180^{\circ}-100^{\circ}=80^{\circ}
\end{aligned}$
Example 3
In the diagram below, O is the centre of the circle. Find the values of α, β.
(a)
$\begin{aligned}
140^{\circ} &=2 \alpha\quad(\because \text { angles substended by arc } A B C) \\
\alpha &=70^{\circ} \\
\angle B+\alpha&=180^{\circ}\quad(\because \text { opposite angles of cyclic quadrilateral } A B C D)\\
\angle B&= 180^{\circ}-\alpha \\
&=180-70^{\circ} \\
&=110^{\circ} \\
\beta&= \angle B C A\quad(\because \text { arc } A B={ arc } B C)
\end{aligned}$
But $\begin{aligned}
\beta_{+} \angle B+\angle B C A &=180^{\circ} \\
2 \beta+110^{\circ}&=180^{\circ} \\
\beta&=\frac{70^{\circ}}{2}\\
\beta&=35^{\circ}
\end{aligned}$
(b)
$\begin{aligned}
\angle D C B&=90^{\circ}\quad(\because \text { angle substended by diameter} D B)\\
\angle D C B+\alpha&=100^{\circ} \quad(\because \text { exterior angle of cyclic quadrilateral } A D B C )\\
\alpha&=100^{\circ}-90^{\circ}=10^{\circ} \\
\angle A+40^{\circ}+\alpha&=180^{\circ} \\
\angle A&=180^{\circ}-\left(40^{\circ}+10^{\circ}\right)=130^{\circ}\\
\angle A+\beta&=180^{\circ} \quad(\because \text { opposite angles of cyclic quadrilateral } A D B C )\\
\beta&=180^{\circ}-130^{\circ}=50^{\circ}\\
\end{aligned}$
Example 4
In the given figure, find the values of x and y.
DM is the perpendicular bisector of AB.
Therefore the center is on DM.
PN is the perpendicular bisector of CD.
Therefore the center is also on PN.
DM and PN intersect at Q.
Therefore Q is the centre of the given circle.
Join AQ.
$\begin{aligned}
A M &=M B \quad\text { (given) } \\
\therefore A M &=3 \mathrm{~cm} \\
A Q^{2} &=A M^{2}+M Q^{2}=3^{2}+4^{2}=25 \\
A Q &=5 \mathrm{~cm} \\
y &=A Q \quad\text { (radii) } \\
\therefore y &=5 \mathrm{~cm} \\
D N &=N C \quad\text { (given) } \\
\therefore D N &=4 \mathrm{~cm} \\
y^{2} &=x^{2}+D N^{2} \\
x &=\sqrt{y^{2}-D N^{2}} \\
&=\sqrt{5^{2}-4^{2}}\\
&=3 \mathrm{~cm}
\end{aligned}$
Example 5
Given ⊙ O, OE= OF, AE= x+2 and CD= 3x-2.Find the radius of the circle.
$\begin{aligned}
O E &\perp A B \\
\therefore A B &=2 A E=2(x+2)=2 x+4 \\
\text { since } O E &=O F \text {, then } A B=C D \\
2 x+4 &=3 x-2 \\
x &=6 \\
\therefore A E &=6+2=8 \\
\end{aligned}$
Join OA.
$\begin{aligned}
\text { radius }=O A&=\sqrt{O E^{2}+A E^{2}} \\
&=\sqrt{6^{2}+8^{2}}\\
&=10
\end{aligned}$
Example 6
Given ⊙ O, AE= 12, BE= x and CE= 2x. Find radius of the circle.
$\begin{aligned}
O B &\perp C D \\
\therefore E D&=C E=2 x \\
\end{aligned}$
$A B$ and $C D$ intersect at $E$ . \\
$\begin{aligned}
\therefore A E &\cdot E B=C E \cdot E D \\
12 \cdot x&=2 x \cdot 2 x \\
x&=3 \\
A B&=A E+E B=12+x=15 \\
\end{aligned}$
$\begin{aligned}
\therefore \text { radius }=O A&=\dfrac{1}{2} A B\\
&=\dfrac{1}{2}(15)\\
&=7.5
\end{aligned}$
Example 7
Given ⊙ O, radius= 6, PC= 3, PA= 5 and EF= 6. Find AB and PE.
$\begin{aligned}
&D C=\text { diameter }=12 \\
&P D=P C+C D=3+12=15 \\
&P A \cdot P B=P C \cdot P D \quad(\text { Theorem } 12) \\
&5 \cdot(5+A B)=3 \times 15 \\
&5+A B=9 \\
&A B=4 \\
&P E \cdot P F=P C \cdot P D \quad(\text { Theorem } 12) \\
&P E(P E+6)=3 \times 15 \\
&P E^{2}+6 P E=45 \\
&P E^{2}+6 P E+9=45+9 \\
&(P E+3)^{2}=54 \\
&P E+3=\sqrt{54} \\
\end{aligned}$
$\begin{aligned}
\therefore P E&=\sqrt{54}-3\\
&=3 \sqrt{6}-3 \\
\end{aligned}$
Exercise 9.1
The point O is the centre of the given circle. Find the values of x, y and z.
$\begin{aligned}
\alpha+30^{\circ}+40^{\circ} &=180^{\circ} \\
\alpha&=180^{\circ}-70^{\circ}=110^{\circ} \\
y+\alpha &=180^{\circ}\quad(\because \text { opposite angles of a cyclic quadrilateral }) \\
y &=180^{\circ}-\alpha \\
&=180^{\circ}-110^{\circ}\\
&=70^{\circ} \\
x+\beta &=180^{\circ}\quad(\because\text { opposite angles of a cyclic quadrilateral }) \\
x &=180^{\circ}-\beta \\
&=180^{\circ}-70^{\circ} \\
&=110^{\circ}
\end{aligned}$
(e)
$\begin{aligned}
\alpha&=50^{\circ}\quad(\because\mathrm{radii}) \\
\alpha+\alpha_{1}+50^{\circ} &=180^{\circ} \\
\alpha_{1} &=180^{\circ}-\left(\alpha+50^{\circ}\right) \\
&=180^{\circ}-\left(50^{\circ}+50^{\circ}\right) \\
&=80^{\circ} \\
\beta&=60^{\circ}\quad(\because\mathrm{radii}) \\
\beta+\beta_{1}+60^{\circ} &=180^{\circ} \\
\beta_{1} &=180^{\circ}-\left(\beta+60^{\circ}\right) \\
&=180^{\circ}-\left(60^{\circ}+60^{\circ}\right) \\
&=60^{\circ} \\
\alpha_{1}+\beta_{1}+\theta &=180^{\circ} \\
\theta &=180^{\circ}-\left(\alpha_{1}+\beta_{1}\right) \\
&=180^{\circ}-\left(80^{\circ}+60^{\circ} \right) \\
&=40^{\circ} \\
x &=\gamma \quad(\because\mathrm{radii}) \\
y &=\phi \quad(\because\mathrm{radii}) \\
x+\gamma+\theta &=180^{\circ} \\
2 x &=180^{\circ}-\theta \\
x &=\dfrac{180^{\circ}-40^{\circ}}{2}=10^{\circ}\\
\theta&=y+\phi\quad(\because\text{exterior angle of a triangle})\\
\theta&=2 y \\
y&=\dfrac{\theta}{2}=\dfrac{40^{\circ}}{2}=20^{\circ}
\end{aligned}$
(f)
$\begin{aligned}
y&=110^{\circ}\quad(\because\text{ exterior angle of a cyclic quadrilateral)}\\
\operatorname{arc} A C&=\operatorname{arc} B D\\
\therefore\alpha&=\beta \\
A C&=B D \\
\therefore A B& ∥ C D \\
\therefore y+z&=180^{\circ} \\
z &=180^{\circ}-y\\
z &=180^{\circ}-110^{\circ}\\
&=70^{\circ}\\
x+z&=180^{\circ}\quad (\because \text{ opposite angles of a cyclic quadrilateral)}\\
x&=180^{\circ}-z\\
&=180^{\circ}-70^{\circ}\\
&=110^{\circ}
\end{aligned}$
ABC is an acute triangle inscribed in ⊙O, and OD is the perpendicular drawn O to BC. Prove that ∠ BOD= ∠ BAC.
$\begin{aligned}
\text{ Given }&:\triangle A B C \text{ is inscribed in } \odot O and O D \perp B C.\\
\text{ To prove }&:\angle B O D=\angle B A C.\\
\text{ Proof }&: \text{ Draw } OC.\\
&\text { In } \triangle B O D \text { and } \triangle C O D, \\
&B D=D C \quad(\because O D \perp B C) \\
&O D=O D \quad(\because\text { common side }) \\
&O B=O C \quad (\because\text { radii })\\
&\therefore \triangle B O D \cong \triangle C O D\quad(\because\text{ S S S }) \\
\end{aligned}$
$\begin{aligned}
\therefore \alpha &=\beta\\
\angle B O D &=\dfrac{1}{2} \angle B O C\\
\text {But} \quad \angle B A C &=\dfrac{1}{2} \angle B O C \quad(\because\text{ angles substended by arc B C })\\
\therefore \angle B O D &=\angle B A C
\end{aligned}$
In ⊙O, two chords AB and CD intersect in the circle at P. Show that ∠ APD= 1/2 ( ∠ AOD + ∠ BOC).
$\begin{aligned}
\text{ Given }&: \text{ In }⊙ O, A B \text{ and } C D \text{ intersect at } P.\\
\text { To prove }&: \angle A P D=\dfrac{1}{2}(\angle A O D+\angle B O C)\\
\text { Proof }&:\alpha=\dfrac{1}{2} \angle A O D \quad(\because\text{ angles substended by arc AD })\\
&\quad\beta=\dfrac{1}{2} \angle B O C \quad(\because\text{ angles substended by arc B C } )\\
&\angle A P D=\alpha+\beta \quad\text{(exterior angle of } \triangle A P C )\\
&\quad\quad\quad=\dfrac{1}{2} \angle A O D+\dfrac{1}{2} \angle B O C \\
&\quad\quad\quad=\dfrac{1}{2}(\angle A O D+\angle B O C)
\end{aligned}$
Two circles intersect at M and N. From M, diameters MA, MB are drawn in each circle. If A, B are joined to N, prove that ANB is a straight line.
$\begin{aligned}
\text{ Given }:&\text{ Two circles intersect at } M \text{ and } N. M A \text{ and } M B \text{ are diameters }.\\
\text{ To prove }:& ANB \text{ is a straight line }.\\
\text{ Proof }:&\alpha=90^{\circ}\quad(\because \text{ angle substended by a díameter })\\
&\beta=90^{\circ} \quad(\because \text{ angle substended by a diameter })\\
&\alpha+\beta=90^{\circ}+90^{\circ}=180^{\circ}\\
&\therefore A N B\text{ is a straight line }
\end{aligned}$
OA and OB are two radii of a circle meeting at right angles. From A, B two parallel chords AX, BY are drawn. Prove that AY ⊥ BX.
$\begin{aligned}
\text{ Given } : & \text{ In } ⊙ O, O A ⊥ O B \text{ and } A X ∥ B Y . \\
\text { To prove } :& A Y \perp B X . \\
\text { Proof } :& \theta =\alpha+\beta \quad(\because \text { exterior angle of a triange }) \\
&\quad=\alpha+\gamma \quad(\because \beta= \gamma,\quad A X ∥B Y) \\
&\quad=\dfrac{1}{2} \angle A O B+\dfrac{1}{2} \angle A O B \\
&\quad=\angle A O B \\
&\quad=90^{\circ} \\
&\therefore A Y ⊥B X
\end{aligned}$
Two circles intersect at R and S. Two straight lines ARB and SCD are drawn meeting one circle at A, C and the other at B, D. Prove that AC ∥ BD. If AB ∥ CD, show that AB= CD.
$\begin{aligned}
\text{ Given }:& \text{ Two circles intersect at } R \text{ and } S.\\
& A R B \text{ and } C S D \text{ are two straight lines }.\\
\text{ To prove }:& A C \| B D\\
& A B \| C D \Rightarrow A B=C D\\
\text{ Proof }:&\text{ Draw } R S.\\
&\alpha+\theta =180^{\circ}\quad(\because \text { opposite angles of a cyclic quadrilateral }) \\
&\quad\quad\theta =\beta \quad(\because\text { exterior angle of a cyclic quadrilateral }) \\
&\therefore \alpha+\beta =180^{\circ} \\
&\therefore A C \| B D \\
& A B \| C D \text { (given) } \\
&\therefore A B C D \text { is a parallelogram. } \\
&\therefore A B=C D
\end{aligned}$
Exercise 9.2
In the following figures, O is the centre of circles. Find the values of x and y.
(a)
$\begin{aligned}
O B&=8 \\
\triangle & O B P \text { is a } 30^{\circ}-60^{\circ} \text { right triangle. } \\
\therefore O P&=4, P B=4 \sqrt{3} \\
O Q&=O P\quad(\because\text { given }) \\
\therefore C D&=A B \quad(\because\text { Theorem } 9 ) \\
\dfrac{1}{2} C D&=\dfrac{1}{2} A B\\
C Q&=P B\\
\text {But}\quad C Q&=Q D,A P =P B\quad(\because\text{ Theorem } 7)\\
\therefore x&=P B=4 \sqrt{3} \\
\end{aligned}$
(b)
$\begin{aligned}
A P&=P B \\
\therefore O P &\perp A B \quad(\because\text { Theorem } 6 ) \\
O P&=5 \quad(\because\text { given }) \\
\triangle & O A P \text { is a } 30^{3}-60^{\circ} \text { right triangle }. \\
\therefore A P&=5 \sqrt{3}, O A=10 \\
y&=A P=5 \sqrt{3} \\
\text { In }\text { rt } &\triangle O Q D, \\
O D^{2} &=O Q^{2}+Q D^{2} \\
Q D^{2} &=O D^{2}-O Q^{2} \\
x^{2}&=O D^{2}-O Q^{2} \\
&=10^{2}-8^{2} \\
&=36 \\
x &=6
\end{aligned}$
In ⊙O, chord AB is perpendicular to CD at P, AB=16, CP=4, PD= 10. Find the radius.
$\begin{aligned}
\quad A B&=16, C P=4, P D=10 \\
\text { Join } & O D \text {. } \\
\text { Draw } & O E \perp A B, O F \perp C D \text {. } \\
A E&=E B=\dfrac{16}{2}=8 \quad(\because\text { Theorem } 7 ) \\
C F&=F D=\dfrac{10+4}{2}=7 \quad(\because\text { Theorem 7 } ) \\
O E P F& \text{ is a rectangle }.\\
\text { Let } E P&=O F=x \\
\therefore E B&=8-x, A P=8+x \\
A P \cdot P B&=C P \cdot P D \quad(\because\text { Theorem } 11 ) \\
(8+x)(8-x)&=4 \times 10 \\
64-x^{2}&=40 \\
x^{2}&=64-40=24 \\
O F^{2}&=24 \\
\text { In }\text { rightt } &\triangle O F D,\\
O D^{2} & =O F^{2}+F D^{2} \\
& =24+7^{2} \\
&=24+49 \\
&=73 \\
O D & =\sqrt{73}
\end{aligned}$
In the figure O is the centre of the concenteric circles and ON ⊥ AB. If OC=10, ON= 8 and OB= 17, find AC.
$\begin{aligned}
&O C=10, O N=8, O B=17 \\
&O N \perp A B \\
&\therefore A N=N B\quad(\text { Theorem } 7) \\
\end{aligned}$
$\begin{aligned}
C N^{2} &=O C^{2}-O N^{2} \\
&=10^{2}-8^{2}=100-64=36 \\
C N &=6 \\
N B^{2} &=O B^{2}-O N^{2} \\
&=17^{2}-8^{2}=289-64=225 \\
N B &=15 \\
\therefore A N &=15 \\
A C &=A N-C N \\
&=15-6=9
\end{aligned}$
Prove Theorem 10: Of any twp chords of a circle, the greater chord is nearer to the centre, and conveersely, the chord nearer to the centre is larger.
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙O, O E \perp A B \text{ and } OF \perp C D \text{ with } A B>C D.\\
\text{ To prove } :& OE < O F \\
\text{ Proof } :& \text{ Let } A B=2 a, C D=2 b, O E=x, OF =y \text{ and radius } =r.\\
& r^{2}=a^{2}+x^{2} and r^{2}=b^{2}+y^{2}\\
&\therefore a^{2}+x^{2} =b^{2}+y^{2} \\
& b^{2}-a^{2} =x^{2}-y^{2}\\
&\text{ Since } b < a, b^{2}< a^{2}\\
& b^{2}-a^{2} < 0 \\
& x^{2}-y^{2} < 0 \\
& x^{2} < y^{2} \\
& x < y \\
&\therefore O E < O F
\end{aligned}$
Conversely,
$\begin{aligned}
\text{ Given } :& In ⊙ O, O E \perp A B \text{ and } O F \perp C D \text{ with } O E<O F.\\
\text{ To prove } :& A B > C D \\
\text{ Proof } :& \text{ Let } A B=2 a, C D=2 b, O E=x,\\
& O F=y \text{ and radius } =r.\\
& r^{2}=a^{2}+x^{2} \text{ and } r^{2}=b^{2}+y^{2}\\
& \therefore a^{2}+x^{2}=b^{2}+y^{2}\\
& x^{2}-y^{2}=b^{2}-a^{2}\\
&\text{ Since } x < y, x^{2} < y^{2}\\
& x^{2}-y^{2} < 0 \\
& b^{2}-a^{2} < 0 \\
& b^{2} < a^{2} \\
& b < a \\
&\therefore a >b \\
&\therefore \quad A B >C D
\end{aligned}$
Let P be a point inside a circle.AB is the diameter passes through P and CPD is the chord perpendicular to AB. Show that CD is the shortest of all chords passing through P.
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙ O, A B \text{ is a diameter }.\\
&C D \perp A B \text{ and }X Y \text{ is any line passing through } P.\\
\text{ To prove } :& C D<X Y\\
\text{ Proof } :& \text{ Draw } O Q \perp X Y.\\
&\text{ In right } \triangle O Q P,\\
& O Q < O P\\
& \therefore X Y\rangle C D \quad(\because\text{ Theorem }10)\\
& A B \text{ is the longest chord }.\\
& \therefore C D<X Y<A B.\\
& \therefore C D \text{ is the shortest of all chords passing through } P.
\end{aligned}$
Through a point P in a circle, the longest chord that can be drawn is 10cm long and the shortest chord is 6cm long. What is the radius of the circle and how far is P from the centre?
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙ O, A B \text{ is the longest chord and } C D \text{ is the shortest chord }.\\ &\text{ The two chords intersest at } P\text{ and } O P \perp C D. \\
\text{ To find } :& O C, O P \\
\text{ Solution } :&\text{ Since the longest chord is the diameter },\\
& A B=10 \mathrm{~cm} \\
&\text { radius }=O C=5 \mathrm{~cm} \\
& C D=6 \mathrm{~cm} \quad (\because\text { given })\\
&\text{ Draw } C D \perp A B.\\
& C P=P D=\dfrac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm} \\
&\text { Let } O P=x \mathrm{~cm} \\
&\text { AP.PB }=C P \cdot P D\quad(\because\text { Theorem 11 }) \\
&(5-x)(x+5) =3\times3 \\
& 25-x^{2} =9 \\
& x^{2} =16 \\
& x =4 \\
& O P =4 \mathrm{~cm}\\
&\therefore P \text{ is } 4 \mathrm{~cm} \text{ far from the centre }.\\
\end{aligned}$
In ⊙O, chord AB and CD are equal and intersect in the circle at E such that AE< EB and CE < ED. Show that △BDE is isosceles with base BD.
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙ O, A B=C D.\\
& A B \text{ and } C D \text{ intersect at E }\\
&\text{ such that } A E < E B \text{ and } C E<E D\\
\text{ To prove } :& \triangle B D E \text{ is isosceles with base } B D\\
\text{ Proof } :& \text{ Join } O E, O P \perp A B \text{ and } O Q \perp C D.\\
& A P=P B \quad(\because\text { Theorem 7 }) \\
& C Q=Q D \quad(\because\text { Theorem 7 }) \\
& B u t \quad A B=C D \\
&\dfrac{1}{2} A B=\dfrac{1}{2} C D\\
& P B=Q D\\
&\text{ In right } \triangle O P E \text{ and } \triangle O Q E,\\
& O P =O Q \quad(A B=C D) \\
& O E =O E \quad(\because\text { common side }) \\
&\angle O P E =\angle O Q E=90^{\circ} \\
&\therefore \triangle O P E \cong \triangle O Q E\quad(\because\text { RHL }) \\
&\therefore E P =E Q \\
& E P+P B =E Q+Q D \\
& E B =E D\\
&\therefore \triangle B D E \text { is isosceles with base } BD.
\end{aligned}$
In ⊙O,congruent chords AB and CD are produced to meet at P. Prove that △PAC is isosceles .
$\begin{aligned}
\text { Given } :& \text { In } ⊙ O, A B=C D \text {. } \\
\text { To prove } :& \triangle P A C \text { is isosceles. } \\
\text { Proof } :& \text { Join OP }, OE \perp A B, O F \perp C P . \\
& A E=E B \quad(\because\text { Theorem } 7)\\
& C F=F D \quad(\because\text { Theorem } 7) \\
& A B=C D\quad(\because\text {given})\\
&\dfrac{1}{2} A B=\dfrac{1}{2} C D\\
& A E=C F--- \text { (1) } \\
& O E=O F \quad(\because\text { Theorem } 9) \\
&\text { In } \triangle O P E \text { and } \triangle O P F, \\
& O P=O P\quad(\because\text { common side }) \\
& O E=O F \quad(\because\text { proved }) \\
&\angle O E P=\angle O F P=90^{\circ} \\
&\triangle O P E \cong \triangle O P F\quad(\because\text { RHL } ) \\
&\therefore P E=P F --- \text { (2) } \\
&\text { Eqn: (1) }+(2), \\
& P E+A E=P F+C F \\
&\therefore P A=P C \\
&\therefore \triangle P A C \text { is isosceles. }
\end{aligned}$
In ⊙O,AB and BC are equal chords.OV ⊥ AB, and OU ⊥ BC. Prove that B is the midpoint of arc VU.
$\begin{aligned}
\text { Given } :& \text { In } ⊙ O, \quad A B=B C \text {, } \\
& O V \perp A B \text {, } O U \perp B C \text {. } \\
\text { To prove } :& B \text { is the midpoint of } \operatorname{arc} V U \text {. } \\
\text { Proof } :& A B=B C \\
&\dfrac{1}{2} A B=\dfrac{1}{2} B C \\
& B D=B E \\
&\text { Join } O B \text {. } \\
&\text { In } \triangle B D O \text { and } \triangle B E O \text {, } \\
& B D=B E\quad(\because \text { proved ) } \\
& O B=O B \quad\text { ( } \because \text { common side) } \\
&\angle B D O=\angle B E O=90^{\circ} \\
&\triangle B D O \cong \triangle B E O \quad\text { (RHL) } \\
&\therefore \alpha=\beta \\
&\text { arc VB }=\operatorname{arc BU} \\
&\therefore B \text { is the midpoint of } \text { arc VU.}
\end{aligned}$
In parallelogram PQRS, PQ=5cm, PR=8cm, QS=6cm. Calculate the lengths of AR and BR.
$\begin{aligned}
\text{ Given } :& In parallelogram P Q R S,\\
& S R=P Q=5 \mathrm{~cm}, P R=8 \mathrm{~cm}, O S=6 \mathrm{~cm}\\
\text{ To find } :& A R, B R\\
\text{ Solution } :& P C=C R=\dfrac{8 \mathrm{~cm}}{2}=4 \mathrm{~cm} \\
& S C=C Q=\dfrac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm} \\
& P C \cdot C A=S C \cdot C Q \quad(\because\text { Theorem 11 }) \\
& 4 \cdot(4-A R)=3 \times 3 \\
& 16-4 A R=9 \\
& 4 A R=7 \\
& A R=\dfrac{7}{4}=1.75 \mathrm{~cm} \\
& S R \cdot B R=P R \cdot A R \quad(\because\text { Theorem 12 })\\
& 5 \cdot B R=8 \times \dfrac{7}{4} \\
& B R=\dfrac{14}{5}=2.8 \mathrm{~cm}
\end{aligned}$
Chords AB and CD intersect at E and AE=EB. A semicircle is drawn with diameter CD. EF, perpendicular to CD, meets this semicircle at F. Prove that AE=EF.
$\begin{aligned}
\text { Given } :& \text { Chords } A B \text { and } C D \text { intersect at } E \text { and } A E=E B \\
& C F D \text { is a semicircle and } E F \perp C D \\
\text { To prove } :& A E=E F \\
\text { Proof } :&\text { Join } C F \text {, FD. } \\
& \angle C F D=90^{\circ} \quad(\because\text { angle substended by a diameter }) \\
& E F \perp C D \text { and } \angle C F D=90^{\circ}, \\
& E F^{2}=C E \cdot E D \quad(\because\text { Theorem 6 }) \\
& A E \cdot E B=C E \cdot E D \quad(\because\text { Theorem } 11) \\
&\therefore A E^{2}=C E \cdot E D \quad(\because A E=E B) \\
&\therefore A E^{2}=E F^{2} \\
&\therefore A E=E F
\end{aligned}$
⊙O and ⊙P intersect at A and B. Show that OP is the perpendicular bisector of common chord AB.
$\begin{aligned}
\text{ Given } :& ⊙ O \text{ and } ⊙ P \text{ intersect at } A and B.\\
\text{ To prove } :& O P \perp A B \text{ and } A D=B D.\\
\text{ Proof } :& \text{ Join } O A, O B, P A, P B.\\
&\text { In } \triangle O A P \text { and } \triangle O B P, \\
& O A=O B \quad\text { (radii) } \\
& P A=P B \quad\text { (radii) } \\
& O P=O P \quad\text { (common side) } \\
&\therefore \triangle O A P \cong \triangle O B P\quad({S S S}) \\
&\therefore \alpha_{1}=\alpha_{2}, \\
&\text { In } \triangle O A D \text { and } \triangle O B D, \\
& O A=O B \quad\text { (proved) } \\
& O D = O D\quad(\text { common side }) \\
&\alpha_{1}=\alpha_{2} \quad(\text { proved }) \\
&\therefore \triangle O A D \cong \triangle O B D\quad\text({S A S}) \\
&\therefore \angle O D A=\angle O D B \\
&\angle O D A+\angle O D B=180^{\circ} \\
&\therefore \angle O D A=\angle O D B=\dfrac{180^{\circ}}{2}=90^{\circ} \\
&\therefore O P \perp A B \text { and } A D=B D .
\end{aligned}$
.png)
.png)
∠ COD is a central angle substended by arc CD (or chord CD).
∠ PQR is an inscribed angle substended by arc PR (or chord PR).
.png)
$\begin{aligned}
\operatorname{arcPQ}&=\operatorname{arc} Q R=\operatorname{arc} R S\quad(\text { given }) \\
\therefore \angle P O Q&=\angle Q O R=\angle R O S \quad(\text { Theorem } 3) \\
\text { But } \angle P O Q&+\angle Q O R+\angle R O S=90^{\circ} \\
\therefore \angle P O Q&=\angle Q O R=\angle R O S=\frac{90^{\circ}}{3}=30^{\circ} \\
\angle P O R&=\angle P O Q+\angle Q O R=30^{\circ}+30^{\circ}=60^{\circ} \\
\angle P&=\angle R \quad(\because O R=O P \quad\text { radii }) \\
\therefore \angle P&=\angle R=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ} \\
\therefore \triangle &P O R \text { is equalateral. } \\
\therefore P R&=O R \\
\text { But } O R&=O S \quad(\because \text { radii }) \\
\therefore P R&=O S=3 \mathrm{~cm}
\end{aligned}$
.png)
$\begin{aligned}
\angle O A B&=\angle O B A\quad(\because O A= O B \text { radii }) \\
\therefore \angle O A B&=40^{\circ} \\
\alpha&=180^{\circ}-\left(40^{\circ}+40^{\circ}\right)=100^{\circ} \\
\alpha&=2 \beta\quad(\because \text { angles substended by arc } A B) \\
\beta&=\frac{1}{2} \alpha=\frac{1}{2}\left(100^{\circ}\right)=50^{\circ} \\
\theta&=\beta\quad(\because \text { angles substended by } \operatorname{arc} A B) \\
\therefore \theta&=50^{\circ} \\
\end{aligned}$
(b) .png)
$\begin{aligned}
\beta&=\dfrac{1}{2} \times 200^{\circ}\quad(\because \text { angles substended by arc } A C B) \\
\beta&=100^{\circ} \\
\alpha+\beta&=180^{\circ}\quad(\because \text { opposite angles of cyclic quadrilateral } A D B C) \\
\alpha&=180^{\circ}-100^{\circ}=80^{\circ}
\end{aligned}$
.png)
$\begin{aligned}
140^{\circ} &=2 \alpha\quad(\because \text { angles substended by arc } A B C) \\
\alpha &=70^{\circ} \\
\angle B+\alpha&=180^{\circ}\quad(\because \text { opposite angles of cyclic quadrilateral } A B C D)\\
\angle B&= 180^{\circ}-\alpha \\
&=180-70^{\circ} \\
&=110^{\circ} \\
\beta&= \angle B C A\quad(\because \text { arc } A B={ arc } B C)
\end{aligned}$
But $\begin{aligned}
\beta_{+} \angle B+\angle B C A &=180^{\circ} \\
2 \beta+110^{\circ}&=180^{\circ} \\
\beta&=\frac{70^{\circ}}{2}\\
\beta&=35^{\circ}
\end{aligned}$
(b) .png)
$\begin{aligned}
\angle D C B&=90^{\circ}\quad(\because \text { angle substended by diameter} D B)\\
\angle D C B+\alpha&=100^{\circ} \quad(\because \text { exterior angle of cyclic quadrilateral } A D B C )\\
\alpha&=100^{\circ}-90^{\circ}=10^{\circ} \\
\angle A+40^{\circ}+\alpha&=180^{\circ} \\
\angle A&=180^{\circ}-\left(40^{\circ}+10^{\circ}\right)=130^{\circ}\\
\angle A+\beta&=180^{\circ} \quad(\because \text { opposite angles of cyclic quadrilateral } A D B C )\\
\beta&=180^{\circ}-130^{\circ}=50^{\circ}\\
\end{aligned}$
DM is the perpendicular bisector of AB.
Therefore the center is on DM.
PN is the perpendicular bisector of CD.
Therefore the center is also on PN.
DM and PN intersect at Q.
Therefore Q is the centre of the given circle.
Join AQ.
$\begin{aligned}
A M &=M B \quad\text { (given) } \\
\therefore A M &=3 \mathrm{~cm} \\
A Q^{2} &=A M^{2}+M Q^{2}=3^{2}+4^{2}=25 \\
A Q &=5 \mathrm{~cm} \\
y &=A Q \quad\text { (radii) } \\
\therefore y &=5 \mathrm{~cm} \\
D N &=N C \quad\text { (given) } \\
\therefore D N &=4 \mathrm{~cm} \\
y^{2} &=x^{2}+D N^{2} \\
x &=\sqrt{y^{2}-D N^{2}} \\
&=\sqrt{5^{2}-4^{2}}\\
&=3 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
O E &\perp A B \\
\therefore A B &=2 A E=2(x+2)=2 x+4 \\
\text { since } O E &=O F \text {, then } A B=C D \\
2 x+4 &=3 x-2 \\
x &=6 \\
\therefore A E &=6+2=8 \\
\end{aligned}$
Join OA.
$\begin{aligned}
\text { radius }=O A&=\sqrt{O E^{2}+A E^{2}} \\
&=\sqrt{6^{2}+8^{2}}\\
&=10
\end{aligned}$
$\begin{aligned}
O B &\perp C D \\
\therefore E D&=C E=2 x \\
\end{aligned}$
$A B$ and $C D$ intersect at $E$ . \\
$\begin{aligned}
\therefore A E &\cdot E B=C E \cdot E D \\
12 \cdot x&=2 x \cdot 2 x \\
x&=3 \\
A B&=A E+E B=12+x=15 \\
\end{aligned}$
$\begin{aligned}
\therefore \text { radius }=O A&=\dfrac{1}{2} A B\\
&=\dfrac{1}{2}(15)\\
&=7.5
\end{aligned}$
$\begin{aligned}
&D C=\text { diameter }=12 \\
&P D=P C+C D=3+12=15 \\
&P A \cdot P B=P C \cdot P D \quad(\text { Theorem } 12) \\
&5 \cdot(5+A B)=3 \times 15 \\
&5+A B=9 \\
&A B=4 \\
&P E \cdot P F=P C \cdot P D \quad(\text { Theorem } 12) \\
&P E(P E+6)=3 \times 15 \\
&P E^{2}+6 P E=45 \\
&P E^{2}+6 P E+9=45+9 \\
&(P E+3)^{2}=54 \\
&P E+3=\sqrt{54} \\
\end{aligned}$(b)(c)1.png)
(e)(f)1.png)
.png)
$\begin{aligned}
x=70^{\circ} &\quad(\because\text { Corollary 1.1 }) \\
\alpha+70^{\circ} &=180^{\circ}\quad(\because\text { Theorem 2) }\\
\alpha &=180^{\circ}-70^{\circ}=110^{\circ} \\
\beta &=30^{\circ} \quad(\because\text { alternative angles) } \\
\alpha+\beta+y &=180^{\circ} \\
y &=180^{\circ}-\left(110^{\circ}+30^{\circ}\right) \\
&=40^{\circ} \\
\end{aligned}$
(2).png)
$\begin{aligned}
x=& \alpha \\
x+\alpha &=150^{\circ}\quad(\because\text { exterior angle of a triangle }) \\
2 x &=150^{\circ} \\
x &=\dfrac{150^{\circ}}{2}=75^{\circ} \\
y &=\beta \\
y+\beta+\theta &=180^{\circ} \\
2 y+\theta &=180^{\circ} \\
\text{ But } x+\theta &=180^{\circ}\quad(\because\text { Theorem } 2) \\
2 y+\theta &=x+\theta \\
2 y &=x \\
y &=\dfrac{x}{2}=\dfrac{75^{\circ}}{2}=37.5^{\circ}
\end{aligned}$
.png)
$\begin{aligned}
\alpha &=55^{\circ}(\text { radii }) \\
240^{\circ} &=2(\alpha+\beta)\quad(\because\text { Theorem } 1) \\
\alpha+\beta &=120^{\circ} \\
\beta &=120^{\circ}-\alpha \\
&=120^{\circ}-55^{\circ}\\
&=65^{\circ} \\
y &=\beta \quad \text { (radii) } \\
\therefore y &=65^{\circ} \\
x + \alpha+\beta &=180^{\circ}\quad (\because \text { opposite angles of acyclic quadrilateral )} \\
x &=180^{\circ}-(\alpha+\beta) \\
&=180^{\circ}-\left(55^{\circ}+65^{\circ}\right) \\
&=60^{\circ}
\end{aligned}$
.png)
$\begin{aligned}
\alpha+30^{\circ}+40^{\circ} &=180^{\circ} \\
\alpha&=180^{\circ}-70^{\circ}=110^{\circ} \\
y+\alpha &=180^{\circ}\quad(\because \text { opposite angles of a cyclic quadrilateral }) \\
y &=180^{\circ}-\alpha \\
&=180^{\circ}-110^{\circ}\\
&=70^{\circ} \\
x+\beta &=180^{\circ}\quad(\because\text { opposite angles of a cyclic quadrilateral }) \\
x &=180^{\circ}-\beta \\
&=180^{\circ}-70^{\circ} \\
&=110^{\circ}
\end{aligned}$
.png)
$\begin{aligned}
\alpha&=50^{\circ}\quad(\because\mathrm{radii}) \\
\alpha+\alpha_{1}+50^{\circ} &=180^{\circ} \\
\alpha_{1} &=180^{\circ}-\left(\alpha+50^{\circ}\right) \\
&=180^{\circ}-\left(50^{\circ}+50^{\circ}\right) \\
&=80^{\circ} \\
\beta&=60^{\circ}\quad(\because\mathrm{radii}) \\
\beta+\beta_{1}+60^{\circ} &=180^{\circ} \\
\beta_{1} &=180^{\circ}-\left(\beta+60^{\circ}\right) \\
&=180^{\circ}-\left(60^{\circ}+60^{\circ}\right) \\
&=60^{\circ} \\
\alpha_{1}+\beta_{1}+\theta &=180^{\circ} \\
\theta &=180^{\circ}-\left(\alpha_{1}+\beta_{1}\right) \\
&=180^{\circ}-\left(80^{\circ}+60^{\circ} \right) \\
&=40^{\circ} \\
x &=\gamma \quad(\because\mathrm{radii}) \\
y &=\phi \quad(\because\mathrm{radii}) \\
x+\gamma+\theta &=180^{\circ} \\
2 x &=180^{\circ}-\theta \\
x &=\dfrac{180^{\circ}-40^{\circ}}{2}=10^{\circ}\\
\theta&=y+\phi\quad(\because\text{exterior angle of a triangle})\\
\theta&=2 y \\
y&=\dfrac{\theta}{2}=\dfrac{40^{\circ}}{2}=20^{\circ}
\end{aligned}$
(1).png)
$\begin{aligned}
y&=110^{\circ}\quad(\because\text{ exterior angle of a cyclic quadrilateral)}\\
\operatorname{arc} A C&=\operatorname{arc} B D\\
\therefore\alpha&=\beta \\
A C&=B D \\
\therefore A B& ∥ C D \\
\therefore y+z&=180^{\circ} \\
z &=180^{\circ}-y\\
z &=180^{\circ}-110^{\circ}\\
&=70^{\circ}\\
x+z&=180^{\circ}\quad (\because \text{ opposite angles of a cyclic quadrilateral)}\\
x&=180^{\circ}-z\\
&=180^{\circ}-70^{\circ}\\
&=110^{\circ}
\end{aligned}$
$\begin{aligned}
\text{ Given }&:\triangle A B C \text{ is inscribed in } \odot O and O D \perp B C.\\
\text{ To prove }&:\angle B O D=\angle B A C.\\
\text{ Proof }&: \text{ Draw } OC.\\
&\text { In } \triangle B O D \text { and } \triangle C O D, \\
&B D=D C \quad(\because O D \perp B C) \\
&O D=O D \quad(\because\text { common side }) \\
&O B=O C \quad (\because\text { radii })\\
&\therefore \triangle B O D \cong \triangle C O D\quad(\because\text{ S S S }) \\
\end{aligned}$
$\begin{aligned}
\therefore \alpha &=\beta\\
\angle B O D &=\dfrac{1}{2} \angle B O C\\
\text {But} \quad \angle B A C &=\dfrac{1}{2} \angle B O C \quad(\because\text{ angles substended by arc B C })\\
\therefore \angle B O D &=\angle B A C
\end{aligned}$
$\begin{aligned}
\text{ Given }&: \text{ In }⊙ O, A B \text{ and } C D \text{ intersect at } P.\\
\text { To prove }&: \angle A P D=\dfrac{1}{2}(\angle A O D+\angle B O C)\\
\text { Proof }&:\alpha=\dfrac{1}{2} \angle A O D \quad(\because\text{ angles substended by arc AD })\\
&\quad\beta=\dfrac{1}{2} \angle B O C \quad(\because\text{ angles substended by arc B C } )\\
&\angle A P D=\alpha+\beta \quad\text{(exterior angle of } \triangle A P C )\\
&\quad\quad\quad=\dfrac{1}{2} \angle A O D+\dfrac{1}{2} \angle B O C \\
&\quad\quad\quad=\dfrac{1}{2}(\angle A O D+\angle B O C)
\end{aligned}$
$\begin{aligned}
\text{ Given }:&\text{ Two circles intersect at } M \text{ and } N. M A \text{ and } M B \text{ are diameters }.\\
\text{ To prove }:& ANB \text{ is a straight line }.\\
\text{ Proof }:&\alpha=90^{\circ}\quad(\because \text{ angle substended by a díameter })\\
&\beta=90^{\circ} \quad(\because \text{ angle substended by a diameter })\\
&\alpha+\beta=90^{\circ}+90^{\circ}=180^{\circ}\\
&\therefore A N B\text{ is a straight line }
\end{aligned}$
$\begin{aligned}
\text{ Given } : & \text{ In } ⊙ O, O A ⊥ O B \text{ and } A X ∥ B Y . \\
\text { To prove } :& A Y \perp B X . \\
\text { Proof } :& \theta =\alpha+\beta \quad(\because \text { exterior angle of a triange }) \\
&\quad=\alpha+\gamma \quad(\because \beta= \gamma,\quad A X ∥B Y) \\
&\quad=\dfrac{1}{2} \angle A O B+\dfrac{1}{2} \angle A O B \\
&\quad=\angle A O B \\
&\quad=90^{\circ} \\
&\therefore A Y ⊥B X
\end{aligned}$
$\begin{aligned}
\text{ Given }:& \text{ Two circles intersect at } R \text{ and } S.\\
& A R B \text{ and } C S D \text{ are two straight lines }.\\
\text{ To prove }:& A C \| B D\\
& A B \| C D \Rightarrow A B=C D\\
\text{ Proof }:&\text{ Draw } R S.\\
&\alpha+\theta =180^{\circ}\quad(\because \text { opposite angles of a cyclic quadrilateral }) \\
&\quad\quad\theta =\beta \quad(\because\text { exterior angle of a cyclic quadrilateral }) \\
&\therefore \alpha+\beta =180^{\circ} \\
&\therefore A C \| B D \\
& A B \| C D \text { (given) } \\
&\therefore A B C D \text { is a parallelogram. } \\
&\therefore A B=C D
\end{aligned}$
(a)(b)(c).png)
1(aa).png)
$\begin{aligned}
O B&=8 \\
\triangle & O B P \text { is a } 30^{\circ}-60^{\circ} \text { right triangle. } \\
\therefore O P&=4, P B=4 \sqrt{3} \\
O Q&=O P\quad(\because\text { given }) \\
\therefore C D&=A B \quad(\because\text { Theorem } 9 ) \\
\dfrac{1}{2} C D&=\dfrac{1}{2} A B\\
C Q&=P B\\
\text {But}\quad C Q&=Q D,A P =P B\quad(\because\text{ Theorem } 7)\\
\therefore x&=P B=4 \sqrt{3} \\
\end{aligned}$
1(ba).png)
$\begin{aligned}
A P&=P B \\
\therefore O P &\perp A B \quad(\because\text { Theorem } 6 ) \\
O P&=5 \quad(\because\text { given }) \\
\triangle & O A P \text { is a } 30^{3}-60^{\circ} \text { right triangle }. \\
\therefore A P&=5 \sqrt{3}, O A=10 \\
y&=A P=5 \sqrt{3} \\
\text { In }\text { rt } &\triangle O Q D, \\
O D^{2} &=O Q^{2}+Q D^{2} \\
Q D^{2} &=O D^{2}-O Q^{2} \\
x^{2}&=O D^{2}-O Q^{2} \\
&=10^{2}-8^{2} \\
&=36 \\
x &=6
\end{aligned}$
1(c).png)
$\begin{aligned}
(2 x+1+3) \cdot 3&=(13+x-1) \cdot(x-1) \quad(\because\text { Theorem } 12 ) \\
(2 x+4) \cdot 3&=(12+x)(x-1) \\
6 x+12&=12 x-12+x^{2}-x \\
x^{2}+5 x-24&=0 \\
(x+8)(x-3)&=0 \\
x+8&=0 \text ({ or }) x-3=0 \\
x&=-8 \text ({ or }) x=3 \\
x&=-8 \text { is impossible }. \\
\therefore x&=3
\end{aligned}$
2(q).png)
2(a).png)
$\begin{aligned}
\quad A B&=16, C P=4, P D=10 \\
\text { Join } & O D \text {. } \\
\text { Draw } & O E \perp A B, O F \perp C D \text {. } \\
A E&=E B=\dfrac{16}{2}=8 \quad(\because\text { Theorem } 7 ) \\
C F&=F D=\dfrac{10+4}{2}=7 \quad(\because\text { Theorem 7 } ) \\
O E P F& \text{ is a rectangle }.\\
\text { Let } E P&=O F=x \\
\therefore E B&=8-x, A P=8+x \\
A P \cdot P B&=C P \cdot P D \quad(\because\text { Theorem } 11 ) \\
(8+x)(8-x)&=4 \times 10 \\
64-x^{2}&=40 \\
x^{2}&=64-40=24 \\
O F^{2}&=24 \\
\text { In }\text { rightt } &\triangle O F D,\\
O D^{2} & =O F^{2}+F D^{2} \\
& =24+7^{2} \\
&=24+49 \\
&=73 \\
O D & =\sqrt{73}
\end{aligned}$
3.png)
$\begin{aligned}
&O C=10, O N=8, O B=17 \\
&O N \perp A B \\
&\therefore A N=N B\quad(\text { Theorem } 7) \\
\end{aligned}$
$\begin{aligned}
C N^{2} &=O C^{2}-O N^{2} \\
&=10^{2}-8^{2}=100-64=36 \\
C N &=6 \\
N B^{2} &=O B^{2}-O N^{2} \\
&=17^{2}-8^{2}=289-64=225 \\
N B &=15 \\
\therefore A N &=15 \\
A C &=A N-C N \\
&=15-6=9
\end{aligned}$
4.png)
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙O, O E \perp A B \text{ and } OF \perp C D \text{ with } A B>C D.\\
\text{ To prove } :& OE < O F \\
\text{ Proof } :& \text{ Let } A B=2 a, C D=2 b, O E=x, OF =y \text{ and radius } =r.\\
& r^{2}=a^{2}+x^{2} and r^{2}=b^{2}+y^{2}\\
&\therefore a^{2}+x^{2} =b^{2}+y^{2} \\
& b^{2}-a^{2} =x^{2}-y^{2}\\
&\text{ Since } b < a, b^{2}< a^{2}\\
& b^{2}-a^{2} < 0 \\
& x^{2}-y^{2} < 0 \\
& x^{2} < y^{2} \\
& x < y \\
&\therefore O E < O F
\end{aligned}$
Conversely,
$\begin{aligned}
\text{ Given } :& In ⊙ O, O E \perp A B \text{ and } O F \perp C D \text{ with } O E<O F.\\
\text{ To prove } :& A B > C D \\
\text{ Proof } :& \text{ Let } A B=2 a, C D=2 b, O E=x,\\
& O F=y \text{ and radius } =r.\\
& r^{2}=a^{2}+x^{2} \text{ and } r^{2}=b^{2}+y^{2}\\
& \therefore a^{2}+x^{2}=b^{2}+y^{2}\\
& x^{2}-y^{2}=b^{2}-a^{2}\\
&\text{ Since } x < y, x^{2} < y^{2}\\
& x^{2}-y^{2} < 0 \\
& b^{2}-a^{2} < 0 \\
& b^{2} < a^{2} \\
& b < a \\
&\therefore a >b \\
&\therefore \quad A B >C D
\end{aligned}$
5.png)
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙ O, A B \text{ is a diameter }.\\
&C D \perp A B \text{ and }X Y \text{ is any line passing through } P.\\
\text{ To prove } :& C D<X Y\\
\text{ Proof } :& \text{ Draw } O Q \perp X Y.\\
&\text{ In right } \triangle O Q P,\\
& O Q < O P\\
& \therefore X Y\rangle C D \quad(\because\text{ Theorem }10)\\
& A B \text{ is the longest chord }.\\
& \therefore C D<X Y<A B.\\
& \therefore C D \text{ is the shortest of all chords passing through } P.
\end{aligned}$
6.png)
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙ O, A B \text{ is the longest chord and } C D \text{ is the shortest chord }.\\ &\text{ The two chords intersest at } P\text{ and } O P \perp C D. \\
\text{ To find } :& O C, O P \\
\text{ Solution } :&\text{ Since the longest chord is the diameter },\\
& A B=10 \mathrm{~cm} \\
&\text { radius }=O C=5 \mathrm{~cm} \\
& C D=6 \mathrm{~cm} \quad (\because\text { given })\\
&\text{ Draw } C D \perp A B.\\
& C P=P D=\dfrac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm} \\
&\text { Let } O P=x \mathrm{~cm} \\
&\text { AP.PB }=C P \cdot P D\quad(\because\text { Theorem 11 }) \\
&(5-x)(x+5) =3\times3 \\
& 25-x^{2} =9 \\
& x^{2} =16 \\
& x =4 \\
& O P =4 \mathrm{~cm}\\
&\therefore P \text{ is } 4 \mathrm{~cm} \text{ far from the centre }.\\
\end{aligned}$
7.png)
$\begin{aligned}
\text{ Given } :& \text{ In } ⊙ O, A B=C D.\\
& A B \text{ and } C D \text{ intersect at E }\\
&\text{ such that } A E < E B \text{ and } C E<E D\\
\text{ To prove } :& \triangle B D E \text{ is isosceles with base } B D\\
\text{ Proof } :& \text{ Join } O E, O P \perp A B \text{ and } O Q \perp C D.\\
& A P=P B \quad(\because\text { Theorem 7 }) \\
& C Q=Q D \quad(\because\text { Theorem 7 }) \\
& B u t \quad A B=C D \\
&\dfrac{1}{2} A B=\dfrac{1}{2} C D\\
& P B=Q D\\
&\text{ In right } \triangle O P E \text{ and } \triangle O Q E,\\
& O P =O Q \quad(A B=C D) \\
& O E =O E \quad(\because\text { common side }) \\
&\angle O P E =\angle O Q E=90^{\circ} \\
&\therefore \triangle O P E \cong \triangle O Q E\quad(\because\text { RHL }) \\
&\therefore E P =E Q \\
& E P+P B =E Q+Q D \\
& E B =E D\\
&\therefore \triangle B D E \text { is isosceles with base } BD.
\end{aligned}$
8.png)
$\begin{aligned}
\text { Given } :& \text { In } ⊙ O, A B=C D \text {. } \\
\text { To prove } :& \triangle P A C \text { is isosceles. } \\
\text { Proof } :& \text { Join OP }, OE \perp A B, O F \perp C P . \\
& A E=E B \quad(\because\text { Theorem } 7)\\
& C F=F D \quad(\because\text { Theorem } 7) \\
& A B=C D\quad(\because\text {given})\\
&\dfrac{1}{2} A B=\dfrac{1}{2} C D\\
& A E=C F--- \text { (1) } \\
& O E=O F \quad(\because\text { Theorem } 9) \\
&\text { In } \triangle O P E \text { and } \triangle O P F, \\
& O P=O P\quad(\because\text { common side }) \\
& O E=O F \quad(\because\text { proved }) \\
&\angle O E P=\angle O F P=90^{\circ} \\
&\triangle O P E \cong \triangle O P F\quad(\because\text { RHL } ) \\
&\therefore P E=P F --- \text { (2) } \\
&\text { Eqn: (1) }+(2), \\
& P E+A E=P F+C F \\
&\therefore P A=P C \\
&\therefore \triangle P A C \text { is isosceles. }
\end{aligned}$
9q.png)
9(a).png)
$\begin{aligned}
\text { Given } :& \text { In } ⊙ O, \quad A B=B C \text {, } \\
& O V \perp A B \text {, } O U \perp B C \text {. } \\
\text { To prove } :& B \text { is the midpoint of } \operatorname{arc} V U \text {. } \\
\text { Proof } :& A B=B C \\
&\dfrac{1}{2} A B=\dfrac{1}{2} B C \\
& B D=B E \\
&\text { Join } O B \text {. } \\
&\text { In } \triangle B D O \text { and } \triangle B E O \text {, } \\
& B D=B E\quad(\because \text { proved ) } \\
& O B=O B \quad\text { ( } \because \text { common side) } \\
&\angle B D O=\angle B E O=90^{\circ} \\
&\triangle B D O \cong \triangle B E O \quad\text { (RHL) } \\
&\therefore \alpha=\beta \\
&\text { arc VB }=\operatorname{arc BU} \\
&\therefore B \text { is the midpoint of } \text { arc VU.}
\end{aligned}$
10.png)
$\begin{aligned}
\text{ Given } :& In parallelogram P Q R S,\\
& S R=P Q=5 \mathrm{~cm}, P R=8 \mathrm{~cm}, O S=6 \mathrm{~cm}\\
\text{ To find } :& A R, B R\\
\text{ Solution } :& P C=C R=\dfrac{8 \mathrm{~cm}}{2}=4 \mathrm{~cm} \\
& S C=C Q=\dfrac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm} \\
& P C \cdot C A=S C \cdot C Q \quad(\because\text { Theorem 11 }) \\
& 4 \cdot(4-A R)=3 \times 3 \\
& 16-4 A R=9 \\
& 4 A R=7 \\
& A R=\dfrac{7}{4}=1.75 \mathrm{~cm} \\
& S R \cdot B R=P R \cdot A R \quad(\because\text { Theorem 12 })\\
& 5 \cdot B R=8 \times \dfrac{7}{4} \\
& B R=\dfrac{14}{5}=2.8 \mathrm{~cm}
\end{aligned}$
11.png)
$\begin{aligned}
\text { Given } :& \text { Chords } A B \text { and } C D \text { intersect at } E \text { and } A E=E B \\
& C F D \text { is a semicircle and } E F \perp C D \\
\text { To prove } :& A E=E F \\
\text { Proof } :&\text { Join } C F \text {, FD. } \\
& \angle C F D=90^{\circ} \quad(\because\text { angle substended by a diameter }) \\
& E F \perp C D \text { and } \angle C F D=90^{\circ}, \\
& E F^{2}=C E \cdot E D \quad(\because\text { Theorem 6 }) \\
& A E \cdot E B=C E \cdot E D \quad(\because\text { Theorem } 11) \\
&\therefore A E^{2}=C E \cdot E D \quad(\because A E=E B) \\
&\therefore A E^{2}=E F^{2} \\
&\therefore A E=E F
\end{aligned}$
12.png)
$\begin{aligned}
\text{ Given } :& ⊙ O \text{ and } ⊙ P \text{ intersect at } A and B.\\
\text{ To prove } :& O P \perp A B \text{ and } A D=B D.\\
\text{ Proof } :& \text{ Join } O A, O B, P A, P B.\\
&\text { In } \triangle O A P \text { and } \triangle O B P, \\
& O A=O B \quad\text { (radii) } \\
& P A=P B \quad\text { (radii) } \\
& O P=O P \quad\text { (common side) } \\
&\therefore \triangle O A P \cong \triangle O B P\quad({S S S}) \\
&\therefore \alpha_{1}=\alpha_{2}, \\
&\text { In } \triangle O A D \text { and } \triangle O B D, \\
& O A=O B \quad\text { (proved) } \\
& O D = O D\quad(\text { common side }) \\
&\alpha_{1}=\alpha_{2} \quad(\text { proved }) \\
&\therefore \triangle O A D \cong \triangle O B D\quad\text({S A S}) \\
&\therefore \angle O D A=\angle O D B \\
&\angle O D A+\angle O D B=180^{\circ} \\
&\therefore \angle O D A=\angle O D B=\dfrac{180^{\circ}}{2}=90^{\circ} \\
&\therefore O P \perp A B \text { and } A D=B D .
\end{aligned}$
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