(i) The centre of circle lies in interior of the circle.
(ii) A point distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and the chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

40°, 40°

\(\angle\)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°

100°

6 cm

8 cm

\(\angle ABC=180°-\angle ADC=180°-110°=70°\) 
\(\angle ACB=90°\) 
\(\therefore \angle BAC=20°\)
\(\angle ACD=\angle BAC\)   | Alternate interior angles

Equal chords subtend equal angles at the centre.

See a circle

Definition of diameter

\(\angle\)FEH =\(\angle\)AED = 180⁰-\(\left( \frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle B \right) \)

\(\angle\)FGH =\(\angle\)CGB
= 180⁰-\(\left( \frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle B \right) \)
Adding,
\(\angle\)FEH +\(\angle\)FGH = 180⁰
\(\therefore\)  EFGH is a cyclic quadrilateral.

Given: AB = AC

To prove: \(\angle\)ECB + \(\angle\)EDB = 180°
Proof: AB = AC \(\Rightarrow\) \(\angle\)1 = \(\angle\)2
AD=AE\(\Rightarrow\)\(\angle\)3=\(\angle\)4 
\(\angle\)A + \(\angle\)1 + \(\angle\)2 = 180° = \(\angle\)A + \(\angle\)3 + \(\angle\)4
2\(\angle\)1 = 2\(\angle\)3
\(\angle\)1 = \(\angle\)3 = \(\angle\)2 = \(\angle\)4
\(\Rightarrow\) DE II BC
(Corresponding angle)
\(\angle\)1 + \(\angle\)BDE = '180°
(\(\angle\)BDE = \(\angle\)CED, as \(\angle\)3 = \(\angle\)4)
\(\angle\)1 + \(\angle\)CEP = 180°
\(\therefore\) B, C, E, Dare concyclic. Proved.

Construction: Draw RM II SP meeting PQ M.
Proof: PQ||SR     (Given)
PS||MR   (Construction)

\(\therefore\)PMRS is a parallelogram
\(\therefore\) \(\angle\)S = \(\angle\)PMR 
In \(\triangle\)RMQ, RM = RQ (as RM = SP, opp. sides of a parallelogram and PS = QR)
\(\therefore\) \(\angle\)RMQ=\(\angle\)Q 
But \(\angle\)RMQ = 180° - \(\angle\)PMR (linear pair)
= 180°- \(\angle\)S. (proved) 
\(\therefore\) \(\angle\)Q = 180°- \(\angle\)S
Hence \(\angle\)Q + \(\angle\)S = 180°
Since one pair of opposite angles is supplementary, PQRS is cyclic. Proved.

Proof: \(\angle\)BAC = \(\angle\)BDC = 30°
(angles in the same segment) 
In \(\triangle\)DBC,
\(\angle\)BDC + \(\angle\)DBC + \(\angle\)BCD = 180° (A.S.P.)
30° + 70° + \(\angle\)BCD = 180°
\(\angle\)BCD = 80°

Now, AB = BC
\(\therefore\) \(\angle\)BAC = \(\angle\)BCA
(angles opp. to equal sides)
\(\therefore\) \(\angle\)BCA = 30°
\(\Rightarrow\) ECD = 80⁰- 30⁰ = 50⁰