In \(\Delta\)EDC, \(\angle\)EDC + \(\angle\)ECD = \(\angle\)BEC (exterior angle of a \(\Delta\) is equal to the sum of two opposite angles)
\(\Rightarrow\) \(\angle\)EDC + 20° = 130°
\(\Rightarrow\) \(\angle\)EDC = 110° or \(\angle\)BDC =110°
\(\angle\)BAC = \(\angle\)BDC = 110°. (Angle in the same segment)

Given AB and CD are the chords of a circle with centre at O such that AB = CD

To Prove: \(\angle\)AOB = \(\angle\)COD
Proof: In \(\Delta\)AOB and \(\Delta\)COD
AO = CO        (radii of same circle)
AB = CD         (given)
BO = DO        (radii of same circle)
\(\Delta\)AOB\(\cong \) \(\Delta\)COD   (SSS)
\(\angle\)AOB = \(\angle\)COD  (c.p.c.t.) 2 Hence Proved.

50°

25°, 40°, 115°, 30°

\(\angle\)ACB=1/2 * \(\angle\)AOB
=1/2*90^o
=45^o
In \(\Delta\)ACB, \(\angle\)CAB=180^o-(30^o+45^o)
=105^o
\(\angle\)OAB=\(\angle\)OBA
=45^o
(Angles opp. to equal sides of triangle are equal as OA = OB radius of same circle)
\(\angle\)CAO = 105°- \(\angle\)OAB
= 105°-45°
= 60°

The perpendicular from the center bisects the chord.

\(\Rightarrow\) DN =1/2 CD = 6cm
and BM = x,
OD =OB      (Radius of the same circle)
OD² = OB²
\(\Rightarrow\) ON² + ND² = OM² + MB²
\(\Rightarrow\) 8² + 6² = 6² + x²
\(\Rightarrow\) x = 8 cm
\(\Rightarrow\) BM =8
\(\Rightarrow\) AB = 2BM
= 2x8
\(\Rightarrow\) AB = 16 cm.

\(120°\)

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

 ​​​​​\(\angle CDB=\angle CAB\) | 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) 
\(\Rightarrow 40°+\angle ECD=70°\) 
\(\Rightarrow \angle ECD=70°\) 
\(\Rightarrow \angle ECD=70°-40°\) 
\(\Rightarrow \angle ECD=30°\) 

\(\ \angle ADC=\frac { 1 }{ 2 } \angle AOC\)
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
\(=\frac { 1 }{ 2 } \left( \angle AOB+\angle BOC \right) \)
\(\\ =\frac { 1 }{ 2 } \left( 60°+30° \right) =\frac { 1 }{ 2 } \left( 90° \right) =45°.\)

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.

Given: AB and CD are equal chords intersecting at 90°
To prove: OPQR is a square
Proof: Since P and R are the mid-point of AB and CD respectively
\(\therefore\) \(\angle\) OPB = \(\angle\)ORD = 90°
\(\Rightarrow\)\(\angle\)OPQ = \(\angle\)ORQ = 90° 
Since equal chords on a circle are equidistant from the centre.
\(\therefore\) OP= OR
Thus in t10PQ and t10RQ, we have
OP=OR
\(\angle\)OPQ= \(\angle\)ORQ
and OQ= OQ
\(\therefore\) \(\triangle OPB\cong \triangle ORQ\)
Thus in quadrilateral OPQR,
We have
OP = OR, PQ = RQ
and \(\angle\)OPQ = \(\angle\)ORQ = 90°
Hence OPQR is a square.

As D, E, C, B are concylic quadrilateral and ㄥEDB+ㄥECB=180 and ㄥDBC+ㄥDEC=180
As ㄥEDB+ㄥADE=180 and ㄥAED+ㄥDEC=180
Hence, we can say that ㄥADE=ㄥACB and ㄥAED=ㄥABC.
As triangle is isosceles, so we can say ㄥADE=ㄥACB=ㄥAED=ㄥABC
So, DE is parallel to BC and AD = AE & DB = EC and DECB is an isosceles trapezium. So, DC and EB will be equal and if they intersect at 0, AO will be the median of the triangle ABC and triangle ADE as well.

Given: Diameter AB and a chord AC have a common end point A. AB = 20 cm and AC = 12 cm.
To determine: OD
Determination: ∵ OD丄AC
∴ \(AD=DC=\frac { 1 }{ 2 } AC=\frac { 1 }{ 2 } \times 12=6\quad cm\)
| ∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
\(OA=OB=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } \times 20=10\quad cm\)
In right triangle ODA,
OA² = OD² + AD² I By Pythagoras Theorem
⇒ (10)²=OD²+(6)²
⇒ OD=8 cm
Hence, AC is 8 cm far from the centre of the circle.

Given: Two chords PQ and RS of a circle are parallel to each other and AB is the perpendicular bisector of PQ.

To Prove: AB bisects RS.
Proof: ∵ AB is the perpendicular bisector
of PQ
∵ AB passes through the centre O.
|∵ The perpendicular bisector of a chord of a circle passes through the centre
∵ \(PQ\parallel RS\)
∵  AB丄RS
∵ AB bisects RS.
| ∵ The perpendicular drawn from the centre of a circle bisects the chord.

Theorem

\(AQ=QB=\frac { 1 }{ 2 } AB=4\quad cm\) 
\(OA^{ 2 }=OQ^{ 2 }+AQ^{ 2 }\) 
\(⇒\ OQ=3cm\) 
\(CP=PD=\frac { 1 }{ 2 } CD=3\ cm\) 
\(OC^{ 2 }=OP^{ 2 }+CP^{ 2 }x\) 
\( ⇒\ OP=4\ cm\) 
\(PQ=OP+OQ=7\ cm\)

\(AO=OD=\frac { 1 }{ 2 } AD=\frac { 15 }{ 2 } \ cm\) 
\( AM=MB=\frac { 1 }{ 2 } AB=15cm\) 
\(OA^{ 2 }=OM^{ 2 }+AM^{ 2 }\)

\(AC=\sqrt { OA^{ 2 }+OC^{ 2 } } \) 
\(\quad =\sqrt { 13^{ 2 }-5^{ 2 } } =12\quad cm\)

\(BM=MC=\frac { 1 }{ 2 } BC=\frac { 1 }{ 2 } (12)=6 \ cm\) 
\(AB=\sqrt { AM^{ 2 }+BM^{ 2 } } =\sqrt { 8^{ 2 }+6^{ 2 } } =10\ cm\)

Equal chords subtend equal angles at the centre.

∵ \(\angle AOB=\angle COD\)
∴ AB=CD=5 cm

Definition of diameter

(i) Let the butter-chords of the biscuit be AB and CD; and the centre of the biscuit be O.
Join each of A, B, C, D to O.

In \(\triangle\)OAB and In \(\triangle\)OAB and \(\triangle\)OCD,
AB = CD (given)
OA = OC (each equal to radius)
OB =OD
(each equal to radius)
\(\therefore \triangle OAB\cong \triangle OCD\) [S.S.S.]
\(\Rightarrow\) \(\angle\)AOB = \(\angle\)COD (c.p.c.t)
Therefore, the butter-chords subtend equal angles at the centre of the biscuit
\(\Rightarrow\) \(\angle\)AOB = \(\angle\)COD (c.p.c.t)
Therefore, the butter-chords subtend equal angles at the centre of the biscuit.
(ii) We are given the length of either chord is greater than the radius and less than diameter of the circle.
Let length of either chord = l, radius = r, and angle subtended by either butter-chord = 0
Two cases arise:
Case I. If I = r
In this case, the chord and the corresponding radius
form an equilateral triangle with side r
\(\therefore\) \(\theta\)=60°
Case II. If l = 2r 
In this case, the butter-chord passes through the centre.
\(\therefore\) \(\theta\) = 180°
Consequently, we arrive at the following inequality :
60° < 8 < 180°
(Asr < 1< 2r)
Thus, the required range is excluding both.
(iii) (a) Congruency of triangles.
(b) c.p.c.t. (Corresponding parts of congruent triangles are equal.)
(c) Equilateral triangle and its angles
(iv) Industrialist, Thoughtfulness, Self-confidence, Rationality.

(i) Given: A circle C(O, r) and chord AB = chord AC. AD is bisector of \(\angle\)CAB.
To Prove: Centre O lies on the bisector of \(\angle\)BAC
Construction Join Be, meeting bisector AD of \(\angle\)BAC, at M.

Proof: In triangles BAM and CAM,
AB=AC    (given)
\(\angle\)BAM= \(\angle\)CAM  (given)
AM=AM   (Common)
\(​​\triangle BAM\cong \triangle CAM\)   (SAS)
\(\Rightarrow\)  BM=CM
and \(\angle\)BMA = \(\angle\)CMA
As \(\angle\)BMA + \(\angle\)CMA = 180⁰ (linear pair)
\(\Rightarrow\) \(\angle\)BMA = \(\angle\)CMA = 90⁰
\(\Rightarrow\) AM is the perpendicular bisector of the chord BC
\(\Rightarrow\) AM passes through the centre O.
[\(\because\) Perpendicular bisector of chord of a circle passes through the centre of the circle]
Hence, the centre of the park lies on the angle bisector of \(\angle\)BAC
(ii) Congruency of triangles by SAS axiom (Geometry)
(iii) Cleanliness and respect for labour.