As AC and BD bisect each other

Therefore, AC and BD are the diagonals of the parallelogram. 
\(\Rightarrow\) ABCD is a parallelogram. 
\(\Rightarrow\) \(\angle\)A = \(\angle\)Cand \(\angle\)B = \(\angle\)D
But, ABCD is a cyclic quadrilateral
\(\therefore\) \(\angle\)A + \(\angle\)C = 180° and \(\angle\)B + \(\angle\)D = 180° 
\(\Rightarrow\) 2\(\angle\)A = 180° and 2\(\angle\)B = 180°
\(\Rightarrow\) \(\angle\)A = 90° and \(\angle\)B = 90°
\(\Rightarrow\) \(\angle\)A = \(\angle\)B = \(\angle\)C = \(\angle\)D = 90°
\(\Rightarrow\) ABCD is a rectangle and diagonals AC and BD are diameters.

Construction: Join AB

\(\angle\)ABD = 90° (angle in a semi-circle)
\(\therefore\) \(\angle\)ABC = 90° (angle in a semi-circle)
\(\angle\)ABD +\(\angle\)ABC = 180°
\(\therefore\) DBC is a line
\(\therefore\) D, B and C are collinear. Hence Proved

Given: ABCD is a cyclic quadrilateral whose diagonals intersect at P. \(\angle DBC=70°\) and \(\angle BAC=30°\).
Required: To find \(\angle BCD\) .
Determination: \(\angle BDC=\angle BAC(=30°)\)
| Angles in the same segment of a circle are equal
Now, in \(\Delta BCD\) ,
\(\angle BCD+\angle BDC+\angle DBC=180°\)
| ∵ The sum of the three angles of a \(\Delta \) is 180°
⇒  \(\angle BCD+30°+70°=180°\)
⇒  \(\angle BCD+100°=180°\)
⇒  \(\angle BCD=180°-100°=80°\).

(i) x = 2 x 35° = 70°
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(ii) \(x=\frac { 1 }{ 2 } \times 110°=55°\)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iii) \(x=\frac { 1 }{ 2 } \times 70°=35°\)
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iv) x = 180° - (90° + 55°) I ∵ Angle in a semi-circle is 90°
= 180° - 145° = 35°
(v) \(x=\frac { 1 }{ 2 } \times \left( 180°-120° \right) \)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle \(=\frac { 1 }{ 2 } \left( 60° \right) =30°\). 

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: A circle with centre O. PQ is a chord of this circle.
OL is the perpendicular drawn to chord PQ from centre O.
To Prove: PL = QL
Construction: Join OP and OQ.

Proof: In \(\Delta OLP\) and 
OP=OQ             I Radii of the same circle
OL = OL            I Common
\(\angle OLP=\angle OLQ\)     I Each = 90°
∴   \(\Delta OLP\cong \Delta OLQ\)    I By RHS congruence criterion
∴ PL=QL      ICPCT

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.

50°, 80°

Two circles intersect at two points Band C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.
To Prove:\(\angle ACP=\angle QCD\)
Proof:      \(\angle ACP=\angle ABP\)           ...(1)
| Angles in the same segment of a circle are equal
\(\angle QCD=\angle QBD\)          ...(2)
| Angles in the same segment of a circle are equal
\(\angle ABP=\angle QBD\)
| Vertically Opposite Angles
From (1), (2) and (3),
\(\angle ACP=\angle QCD\).             

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.

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 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.

25°

100°

OB=OC=BC
∴ \(\angle BOC=60°\)
∴ \(\angle BAC=\frac { 1 }{ 2 } 60°\angle BOC=30°\)

Theorem

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

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

Definition of diameter