Construction: Join RO and OM.
Proof: SR = SM (given)
\(\angle\)ROS = \(\angle\)SOM (equal chords subtend equal angles at the centre)

In \(\Delta\)ROP and \(\Delta\)MOP,OR = OM  (radius)
\(\angle\)ROS = \(\angle\)SOM  (proved above)
OP = OP   (common)
\(\Delta\)ROP \(\cong\) \(\Delta\)MOP  (SAS)
RP = PM

Given: ABC is a triangle and P is a point on the side BC such that AB = AP. AP produced meets the circumcircle of \(\Delta ABC\) at Q.
To Prove: CP=CQ

Proof: In \(\Delta ABP\) and \(\Delta CQP\),
 \(\angle BAP=\angle QCP\)
| Angles in the same segment of a circle are equal
\(\angle ABP=\angle CPQ\)
| Vertically opposite angles
∴ \(\Delta ABP\cong \Delta CPQ\)
| AA criterion of similarity
∴ \(\frac { AB }{ CQ } =\frac { BP }{ QP } =\frac { AP }{ CP } \)
| ∵ Corresponding sides of two similar triangles are proportional
⇒ ​​​​​​\(\frac { AB }{ CQ } =\frac { AP }{ CP } \)
But  AB=AP   | Given
∴ CQ=CP

\(2x=150°\)
| The angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the remaining part of the circle
⇒ \(x=75°\)                              ....(1)
\(x+\angle BCD=180°\)
| ∵ ACD is a straight line
⇒ \(75°+\angle BCD=180°\)
⇒ \(\angle BCD=105°\)
∵ BEDC is a cyclic quadrilateral
∴ \(\angle BCD+\angle BED=180°\)
I Opposite angles of a cyclic quadrilateral are supplementary
⇒ ​​​​​​\(105°+y=180°\)
 \(y=75°\)   ....(2)
 ​​​​\(z=2y\)
 | The angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the remaining part of the circle
⇒ \(z=2\times 75°=150°\)     ...(3)

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: 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: 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.