Join OC, OD and BC.
Triangle ODC is equilateral
Therefore, \(\angle\) COD = 60°
Now, \(\angle\) CBD =\(\frac{1}{2}\) \(\angle\) COD
This gives \(\angle\) CBD = 30°
Again, \(\angle\) ACB = 90° 
So, \(\angle\) BCE = 180° – \(\angle\) ACB = 90°
Which gives \(\angle\) CEB = 90° – 30° = 60°, i.e., \(\angle\) AEB = 60°

∵ OP = OQ
∴ Chord AB and chord CD are equidistant from the centre.
⇒ \(\overline{\mathrm{AB}}=\overline{\mathrm{CD}}\)
⇒ \(\frac{1}{2} \overline{\mathrm{AB}}=\frac{1}{2} \overline{\mathrm{CD}}\)
⇒ \(\overline{\mathrm{AP}}=\frac{1}{2} \overline{\mathrm{CD}}\)
⇒ 4 cm = \(\frac{1}{2} \overline{\mathrm{CD}}\)
⇒ \(\overline{\mathrm{CD}}\) = 2 x 4 cm
⇒ \(\overline{\mathrm{CD}}\) = 8 cm
Thus, the required length of CD is 8 cm.

Given ED II BC
AB =AC
\(\Rightarrow\) \(\angle\)2 = \(\angle\)3 ...(i)
(Opp. Ls to opp. side are always equal)
In an isosceles II
\(\angle\)1 + \(\angle\)2 = 180° [Interior \(\angle\)s)
\(\therefore\) \(\angle\)3 + \(\angle\)4 = 180⁰
\(\therefore\) \(\angle\)1 + \(\angle\)3 = 180⁰
\(\angle\)2 + \(\angle\)4 = 180° 
but these are opp angles of a quad.
\(\therefore\) BCDE is a cyclic quad.

AE = BE
\(\angle\)A = \(\angle\)B      (angle opposite to equal sides of a \(\Delta\)are equal)
\(\angle\)EDC = \(\angle\)B;
\(\angle\)ECD = \(\angle\)A (exterior angle of cyclic quad.)
\(\therefore\)   \(\angle\)EDC = \(\angle\)A (Corresponding Angles)
\(\therefore\)    AB II DC Proved.

Given Two chordsAB and CD of a circle C (O, r), such that OL < OM, where OL and OM are perpendiculars from O on AB and CD, respectively.
To prove AB > CD
Construction Join OA and OC.
Proof Since, the perpendicular from the centre of a circle to a chord bisects the chord. 
\(\therefore\) AL = 1/2 AB and CM = 1/2 CD
In right angled flOLA and flOMC, we have
OA² = OL² + AL²
and OC² = OM² + CM² [by Pythagoras theorem]
\(\Rightarrow\) AL² = OA² - OL²... (i)
and CM² = OC²- OM²......(ii)
Now, OL \(\Rightarrow\)OL²< OM²
\(\Rightarrow\)- OL²> - OM²
[multiplying by (-) on both sides]
\(\Rightarrow\) OA² - OL²> OA²- OM²
[adding OA2 on both sides]  
\(\Rightarrow\) OA² - OL²> OC²- OM²   [\(\because\)OA²= OC²]
\(\Rightarrow\) AL² > CM² [from Eqs. (i) and (ii)]
\(\Rightarrow\) AL> CM
\(\Rightarrow\) 2AL> 2 CM [multiplying by 2]
\(\Rightarrow\) AB > CD
Hence proved. 

Given Two equal chords AB and CD of a circle intersecting at a point P.
To prove  PB = PD
Construction JoinOP .DrawOL\(\bot\)ABandOM\(\bot\)CD.
Proof Since, equal chords are equidistant from the centre
\(\therefore\) OL = OM [proved above]
In \(\Delta\)OLP and \(\Delta\)OMP,[each 90⁰]
OL = OM[common sides]
\(\angle\)OLP= \(\angle\)OMP
OP = OP
\(\therefore\)\(\Delta\)OLP \(\cong\) \(\Delta\)OMP
[by RHS congruence rule]
\(\Rightarrow\) LP = MP
Now, AB = CD
\(\Rightarrow\) 1/2(AB) = 1/2(CD) 
\(\Rightarrow\) BL = DM
[\(\because\) perpendicular drawn from centre to the circle bisects the chord i.e. AI = LB and CM = MD]
On subtracting Eq. (ii) from Eq. (i), we get
\(\Rightarrow\)LP-BL = MP-DM \(\Rightarrow\) PB = PD(1)

Given: Bisectors of angles A, Band C of a triangle ABC intersect its circumcircle at D, E and F respectively.
To Prove: The angles of the \(\Delta DEF\) are \(90°-\frac { A }{ 2 } ,90°-\frac { B }{ 2 } \) and \(90°-\frac { C }{ 2 } \) respectively.

Proof: \(\angle FDE=\angle FDA+\angle EDA\)
 \(=\angle FCA+\angle EBA\)
| ∵ Angles in the same segment of a circle are equal
 \(=\frac { 1 }{ 2 } \angle C+\frac { 1 }{ 2 } \angle B\)
⇒    \(\angle D=\frac { \angle C+\angle B }{ 2 } \)
\(=\frac { 180°-\angle A }{ 2 } \)
| ∵ In \(\Delta ABC,\quad \angle A+\angle B+\angle C=180°\) (Angle Sum Property)
 \(=90°-\frac { \angle A }{ 2 } \)
Similarly, we can show that
\(E=90°-\frac { \angle B }{ 2 } \)
and  \(F=90°-\frac { \angle C }{ 2 } \).

Given: ABCD is a parallelogram. The circle through A, B and C intersects CD (produced, if necessary) at E.

To Prove: AE = AD.
Proof: In cyclic quadrilateral ABCE,
 \(\angle AED+\angle ABC=180°\)                             ....(1)
| ∵ Opposite angles of a cyclic quadrilateral are supplementary.
Also, \(\angle ADE+\angle ADC=180°\)
I Linear Pair Axiom
But   \(\angle ADC=\angle ABC\)
 Opposite angles of a || gm
∴ \(\angle ADE+\angle ABC=180°\)                                   ....(2)
From (1) and (2), we have
\(\angle AED+\angle ABC=\angle ADE+\angle ABC\)
⇒ \(\angle AED=\angle ADE\)
∴ In triangle ADE,
AE = AD
| ∵ Sides opposite to equal angles of a triangle are equal.
 Hence Proved.

∴ Given: AB DC is a rhombus. E is the point of intersection of its diagonals.

To Prove: The circle drawn with any sides AB of rhombus AB DC as a diameter passes through the point E.
Proof: In \(\Delta AEB\) and \(\Delta AEC\),
AB = AC                | Given
 \(\angle BEA+\angle CEA=180°\)    | Linear Pair Axiom
⇒  \(90°+\angle CEA=180°\)
I Angle in a semi-circle is 90°
⇒  \(\angle CEA=90°\)
∴ \(\angle BEA=\angle CEA=180°\)
AE = AE                       | Common
∴  \(\Delta AEB\cong \Delta AEC\)                      | RHS
∴ BE = CE
⇒ E is the mid-point of BC.
⇒ E is the point of intersection of diagonals.
Hence, the circle drawn with AB as diameter passes through the point E.
Similarly, we can prove that the circle drawn with AC as diameter passes through the point D which is the point of intersection of its diagonals.

Given: ABCD is a cyclic parallelogram.
To Prove: ABCD is a rectangle.

Proof: ABCD is a cyclic quadrilateral
∴ \(\angle 1+\angle 2=180°\)                            ...(1)
| ∵ Opposite angles of a cyclic quadrilateral are supplementary
∵ ABCD is a parallelogram
∴ \(\angle 1=\angle 2\)                                         ...(2)
| Opp. angles of a || gm
From (1) and (2),
 \(\angle 1=\angle 2=90°\)
∴ || gm ABCD is a rectangle.
[A parallelogram with one of its angles 90° is a rectangle]

Given: ABC and ADC are two right triangles with common hypotenuse AC
To Prove:\(\angle CAD=\angle CBD\)

Proof: ∵ AC is the common hypotenuse of two right triangles ABC and ADC.
∴ \(\angle ABC=90°=\angle ADC\)
⇒ Both the triangles are in the same semi-circle.
∴ Points A, B, D and C are concyclic.
∴ DC is a chord
∴ \(\angle CAD=\angle CBD\).
| ∵ Angles in the same segment of a circle are equal

Given: Circles are drawn with sides AB and AC of a triangle ABC as diameters. They intersect at a point D.
To Prove: D lies on the third side BC of \(\Delta \)ABC
Construction: Join AD

Proof: ∵ Circle drawn on AB as diameter intersects BC in D.
∴  \(\angle ADB=90°\)
| Angle in a semi-circle
But \(\angle ADB+\angle ADC=180°\)
Linear Pair Axiom
∴ \(\angle ADC=90°\)
Hence, the circle described on AC as diameter must pass through D.
Thus, the two circles intersect in D.
Now, \(\angle ADB+\angle ADC=180°\).
∴ Points B, D, C are collinear.
∴ D lies on BC.

Given: ABCD is a trapezium whose nonparallel sides AD and BC are equal.
To Prove: Trapezium ABCD is cyclic.
Construction: Draw BE 11 AD.
Proof: ∵ AB || DE        I Given
and         AD || BE       I By construction
∴  Quadrilateral ABCD is a parallelogram.

∴ \(\angle BAD=\angle BED\)                        ....(1)
| Opp.\(\angle \) s of a || gm are equal
and    AD = BE                                 ...(2)
Opp. sides of a || gm are equal
But     AD = BC  .......(3) I Given
From (2) and (3),
BE=BC
∴ \(\angle BEC=\angle BCE\)                         ....(4)
| Angles opposite to equal sides of a triangle are equal
\(\angle BEC+\angle BED=180°\)| Linear Pair Axiom
⇒ \(\angle BCE+\angle BAD=180°\) I From (4) and (1)
⇒ Trapezium ABCD is cyclic.
∵ If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Given: ABCD is a trapezium whose nonparallel sides AD and BC are equal.
To Prove: Trapezium ABCD is cyclic.
Construction: Draw BE 11 AD.
Proof: ∵ AB || DE        I Given
and     AD || BE           I By construction
∴  Quadrilateral ABCD is a parallelogram.

∴ \(\angle BAD=\angle BED\)                        ....(1)
| Opp.\(\angle \) s of a || gm are equal
and    AD = BE                                 ...(2)
Opp. sides of a || gm are equal
But     AD = BC            ...(3) I Given
From (2) and (3),
BE = BC
∴ \(\angle BEC=\angle BCE\)        ....(4)
| Angles opposite to equal sides of a triangle are equal
\(\angle BEC+\angle BED=180°\)       | Linear Pair Axiom
⇒ \(\angle BCE+\angle BAD=180°\) I From (4) and (1)
⇒ Trapezium ABCD is cyclic.
I ∵ If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic

In \(\Delta OAB\) and \(\Delta OCD\),
OA = OC       | Radii of a circle
OB = OD     | Radii of a circle
\(\angle AOB=\angle COD\)        | Vertically Opposite Angles

∴ \(\Delta OAB\cong \Delta OCD\)   | SAS Rule
∴  AB = CD            I CPCT
⇒ Arc AB = Arc CD                             ....(1)
Similarly, we can show that
Arc AD = Arc CB
Adding (1) and (2), we get
Arc AB + Arc AD = Arc CD + Arc CB  ...(2)
⇒  Arc BAD = Arc BCD
⇒  BD divides the circle into two equal parts (each a semicircle)
∴ \(\angle A=90°,\ \angle C=90°\)        Angle in semi-circle is 90°
Similarly, we can show that
\(\angle B=90°,\ \angle D=90°\)
∴ \(\angle A=\angle B=\angle C=\angle D=90°\)
∴  ABCD is a rectangle.

Let BD = x m

Then in right triangle ODB,
OB² = OD² + BD²
By Pythagoras Theorem
⇒ (20)² = OD² + x²
⇒ OD²=  400 - x²⇒ OD =\(\sqrt { 400-x^{ 2 } } \)
Again, area of equilateral triangle ABC
= Area of \(\Delta OBC\) + Area of \(\Delta OCA\) + Area of \(\Delta OAB\)
= 3 Area of \(\Delta OBC\)=  3\(\frac { \left( BC \right) \left( OD \right) }{ 2 } \)
\(=3x\sqrt { 400-x^{ 2 } } \)                       ....(2)
⇒  \(\sqrt { 3 } \sqrt { 400-x^{ 2 } } =x\)
Squaring both sides,
3(400 - x²) = x²
⇒ 1200 - 3x² = x²
⇒  4x² = 1200   ⇒ x² = 300
⇒ \(x=10\sqrt { 3 } \)  ⇒ BD=\(10\sqrt { 3 } \)
⇒  \(2BD=20\sqrt { 3 } \)  ⇒ \(BC=20\sqrt { 3 } \)
Hence, the length of string of each phone is \(20\sqrt { 3 } \) m.

Construction: Draw OL 丄 RS.
Let KR = xm
\(ar\left( \Delta ORS \right) =ar\left( \Delta ORK \right) +ar\left( \Delta SRK \right) \)
\(=\frac { \left( OK \right) \left( KR \right) }{ 2 } +\frac { \left( KS \right) \left( KR \right) }{ 2 }\)
\( \\ =\frac { \left( KR \right) \left( OK+KS \right) }{ 2 } =\frac { \left( KR \right) \left( OS \right) }{ 2 } \)
\(=\frac { \left( x \right) \left( 5 \right) }{ 2 } \ ...........(1)\)

Again, ar \(\left( \Delta ORS \right) \)
\(=\frac { RS\times OL }{ 2 } =\frac { 6\times OL }{ 3 } \)
\(=\frac { 6\times \sqrt { OR^{ 2 }-RL^{ 2 } } }{ 2 } \)                I By Pythagoras Theorem
\(=\frac { 6\times \sqrt { 25-9 } }{ 2 } =\frac { 6\times 4 }{ 2 } =12\quad m\)    .........(2)
From equations (1) and (2),
\(\frac { \left( x \right) \left( 5 \right) }{ 2 } =12\Rightarrow x=\frac { 12\times 2 }{ 5 } =\frac { 24 }{ 5 } =4.8\quad m\)
⇒ KR = 4.8 m
∴ RM = 2KR = 2 x (4.8) = 9.6 m
Hence, the distance between Reshma and Mandip is 9.6 m.

Given: Two equal chords AB and CD of a circle with centre 0 intersect within the circle. Their point of intersection is E.
To Prove: \(\angle OEA=\angle OED\) .

Construction: Join OA and OD.
Proof: In \(\Delta OEA\) and \(\Delta OED\) ,
OE = OE  I Common
OA = OD   I Radii of a circle
AE = DE
| Proved in Example 2 above
∴\(\Delta OEA\cong \Delta OED\)  I SSS Rule
∴ \(\angle OEA=\angle OED.\)     I CPCT

Given: A circle with centre O. Its two equal chords AB and CD intersect at E.
To prove: AE = DE and CE = BE.
Construction: Draw OM 丄 AB and ON 丄 CD join OE.
Proof: In \(\Delta OME\) and \(\Delta ONE\) ,
OM = ON
| ∵ Equal chords of a circle are equidistant from the centre
OE = OE                   I Common

∴\(\Delta OME\cong \Delta ONE\)  | RHS Rule
∴ ME = NE I CPCT
⇒AM + ME = DN + NE
∵ \(AB=CD\Rightarrow \frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } CD\Rightarrow AM=DN\)
⇒  AE = DE
⇒ AB - AE = CD - DE  | ∵ AB=CD
⇒  B E= CE.  | Given

Given: Two circles with centres O and P intersecting at A and B.
Prove: OP is the perpendicular bisector of AB.
Construction: Join OA, OB, PA and PB. Let OP intersect AB at M.
Proof: In \(\Delta \) OAP and \(\Delta \) OBP,
OA = OB         | Radii of a circle
PA = PB      I Radii of a circle
OP = OP     I Common
∴  \(\Delta \ OAP\cong \Delta \ OBP\) I SSS Rule
∴ \(\angle AOP=\angle BOP\)     I CPCT
⇒\(\angle AOM=\angle BOM\)                       ...(1)
In \(\Delta \) AOM and \(\Delta \) BOM,
OA = OB  I Radii of a circle
 \(\angle AOM=\angle BOM\)    | From (1)
OM = OM   I Common
∴\(\Delta AOM\cong \Delta BOM\)    I SAS Rule
∴ AM = BM                  ...(2)
I CPCT
and \(\angle AMO=\angle BMO\)          ... (3)
I CPCT
But ​​​​​​​ \(\angle AMO+\angle BMO=180°\) I Linear Pair Axiom
∴ \(\angle AMO+\angle BMO=90°\)                ...(4)
∴ OM, i.e., OP is the perpendicular bisector of AB. I From (2) and (4)