Given: Bisector AD of  \(\angle BAC\) of  \(\Delta ABC\) passes through the centre O of the circumcircle of \(\Delta ABC\).
To Prove: AB = AC.
Construction: Draw OP丄AB and OQ丄AC.
Proof:

In \(\Delta APO\) and \(\Delta AQO\)
 \(\angle OPA=\angle OQA\)
I Each = 90° (By construction)
 \(\angle OAP=\angle OAQ\)   | Given
OA = OA   | Common
∴ \(\Delta APO\cong \Delta AQO\)   I AAS
∴  OP= OQ  I CPCT
∴ AB = AC.  I ∵ Chords equidistant from the centre of a circle are equal.

Given: A circle with centre O. PQ is a chord of this circle. M is the mid-point of the chord PQ.
To Prove: OM 丄 PQ.
Construction: Join OP and OQ.

Proof: In \(\Delta OMP\) and \(\Delta OMQ\),
OP = OQ    I Radii of the same circle
OM = OM   I Common
MP = MQ
I ∵ M is the mid-point of PQ
∴  \(\Delta OMP\cong \Delta OMQ\)    I By SSS congruence criterion
∴  \(\angle OMP=\angle OMQ\)     I CPCT
But \(\angle OMP+\angle OMQ=180°\) I Linear pair axiom
∴ \(\angle OMP+\angle OMQ=90°\)
⇒   OM 丄 PQ.

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.

Draw OP perpendicular to xy from the centre to a chord bisecting it.
OP \(\bot\) to chord BC
\(\Rightarrow\) BP = PC      ...(i)
Similarly, AP = PD       ..(ii)
Subtracting eqn. (i) from eqn. (ii), we get
AP- BP = PD-PC
or AB = CD

Since sum of the opposite pairs of angles in a cyclic quadrilateral is 180°.
Hence ㄥB+ㄥD=180°
ㄥB=180°-70°
=110°
Again, AB II CD and AD is its transversal, so
ㄥA+ㄥD=180°
ㄥA=180°-70°
=110°
and ㄥA+ㄥC=180°
⇒ 110°+ㄥC=180°
ㄥC=180°-110°
=70°

36, 60

70°

8 cm

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 \(\Delta ABC\),
\(\angle BAC+\angle ABC+\angle ACB=180°\)
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.

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

\(AC=CB=\frac { 1 }{ 2 } AB=4\ cm\) 
\(OA^{ 2 }=OC^{ 2 }+AC^{ 2 }\)
⇒ OC=3 cm
CD=OD-OC=5-3=2 cm

A diameter is the largest chord. Diameter\(=2\times \)Radius.

Equal chords subtend equal angles at the centre.

See a circle

(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.