Construction: Join OP.
Proof: In \(\Delta\)OMP and \(\Delta\)ONP,
\(\angle\)OMP = \(\angle\)ONP = 90°  (given)
OP =OP        (common)
MP =NP    (given)
\(\Delta\)OMP :;\(\Delta\)ONP    (RHS)
\(\therefore\) OM=ON    (c.p.c.t.)
\(\therefore\) AB =CD  (chords equidistant from centre are equal)

Given: OD \(\bot\)BC
Proof: In \(\Delta\)OBD and \(\Delta\)OCD
OB = OC        (radii)
OD = OD       (Common)
\(\angle\)ODB = \(\angle\)ODC     (90°)
\(\Delta\)OBD \(\cong \) \(\angle\)OCD    (RHS rule)
\(\angle\)BOD = \(\angle\)COD    (c.p.c.t.) 1
But \(\angle\)BOC = 2\(\angle\)BOD = 2\(\angle\)BAC
\(\Rightarrow\) \(\angle\)BOD = \(\angle\)BAC

140°

120°

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: 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: 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.

 ​​​​​\(\angle CDB=\angle CAB\) | Angles in the same segment of a circle are equal
=40°   .............(1)
 ​​​\(\angle DBC=70°\)  ..............(2)
In \(\Delta BCD\),

\(\angle BCD+\angle DBC+\angle CDB=180°\) | Sum of all angles of a triangle is 180°
⇒ \(\angle BCD+70°+40°=180°\)| Using (1) and (2)
⇒  \(\angle BCD+110°=180°\)
⇒ ​​​​​​ ​​\(\angle BCD=180°-110°\)
⇒  \(\angle BCD=70°\)   ............(3)
In ​​​​​​​\(\Delta ABC\),
 AB=BC
∴ ​​​​​​​\(\angle BCA=\angle BAC\)| Angles opposite to equal sides of a triangle are equal
 =40° .........(4)
 \(|\because \quad \angle BAC=30°\) (given)
Now,​​​ ​​​​​​​\(\angle BCD=70°\)   | From (3) ​​​​​​​
\(\Rightarrow angle BCA+\angle ECD=70°\) | From (4) 
\(\Rightarrow 40°+\angle ECD=70°\) 
\(\Rightarrow \angle ECD=70°\) 
\(\Rightarrow \angle ECD=70°-40°\) 
\(\Rightarrow \angle ECD=30°\) 

\(\Rightarrow \)\(\angle CED+130°=180°\)
\(\Rightarrow \) \(\angle CED=180°-130°=50°\)  ..........(1)
 \(\angle ECD=20°\)   .............(2)
In \(\Delta CED\),
\(\angle CED+\angle ECD+\angle CDE=180°\) | Sum of all the angles of a triangle is 180°
\(\Rightarrow \) 50°+20°+\(\angle CDE=180°\) | Using (1) and (2)
\(\Rightarrow \) \(70°\angle CDE=180°\)
\(\Rightarrow \) ​​​​​\(\angle CDE=180°\)-70°
\(\Rightarrow \) \(\angle CDE=110°\)     ...........(3)
Now, \(\angle BAC=\angle CDE\) 
| Angles in the same segment of a circle are equal=110°.   |Using (3)

In \(\Delta ABC\),
\(\angle BAC+\angle ABC+\angle ACB=180°\)
Sum of all the angles of a triangle is 180°
⇒ \(\angle BAC+69°+31°=180°\)
⇒ \(\angle BAC+100°=180°\)
⇒ ​​​​​​ \(\angle BAC=180°-100°=80°\)  .........(1)
Now, \(\angle BDC=\angle BAC\)
Angles in the same segment of a circle are equal = 80°.       Using (1)

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.

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

Let O and O' be the centres of circles of radii 5 cm and 3 cm respectively. Let PQ be the common chord of the two circles.
∵ 5² = 4² + 3²
∴ OP² = OO'² + O'P²
⇒ \(\angle OO'P=90°\)
I By Converse of Pythagoras Theorem
⇒ O' lies on the common chord PQ.
∵ OO' 丄 PQ
∴ OO' bisects PQ
I The perpendicular drawn from the centre of a circle to a chord of it bisects the chord PQ.
∵ O' is the mid-point of PQ.
Therefore, length of the common chord
= PQ = 20'P
= 2 x 3 = 6 cm

(i)
 
No point common
(ii)

One point common
(iii)

Two points common
Each pair has at the most two common points. The maximum number of common points is two.