Given: ABCD is a cyclic quadrilateral. AC bisects both the angles A and C.
To Prove: \(\angle ABC=90°\)

Proof: In \(\Delta ADC\) and \(\Delta ABC\) ,
\(\angle DAC=\angle BAC\)  | ∵ AC bisects angle A
\(\angle DCA=\angle BCA\) | ∵ AC bisects angle C
AC=AC    | Common
∴ \(\Delta ADC\cong \Delta ABC\) | ASA congruence rule
∴ \(\angle ADC=\angle ABC\)    | CPCT
But \(\angle ADC+\angle ABC=180°\)
| ∵ Opposite angles of a cyclic quadrilateral are supplementary
∴ \(\angle ADC=\angle ABC=90°\) ​​​

Given: AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E.
To Prove: EB = ED and AE = CE.

Construction: From O draw OP丄 AB and OQ丄CD. Join OE.
Proof: AB = CD
∴  OP=OQ I ∵ Equal chords of a circle are equidistant from the centre
Now in right \(\quad \Delta s\) OPE and OQE,
Hyp. OE = Hyp. OE  I Common
Side OP = Side OQ  I Proved above
∴ \(\Delta OPE\cong \Delta OQE\)
I R.H.S. Congruence Axiom
∴  PE= QE      I CPCT
⇒ \(PE-\frac { 1 }{ 2 } AB=QE-\frac { 1 }{ 2 } CD\)
| ∵ AB = CD (Given)
⇒ PE-PB = QE-QD
⇒ EB = ED.
⇒ BE + AB = ED + CD I ∵ AB = CD     | ∵ AB = CD
⇒ AE=CE

Given: OD is perpendicular to chord AB of a circle where centre is O. BC is a diameter of the circle.
To Prove: CA=20D
Proof: OD丄AB
∴ D is the mid-point of AB
 |The perpendicular drawn from the centre of a circle to a chord bisects the chord.
In \(\Delta BAC\),
∵ \(OD\parallel AC\) I By mid-point theorem
and \(OD=\frac { 1 }{ 2 } AC\)
⇒ CA=2 OD

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°

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.

\(\angle\)AOD + \(\angle\)BOD = 180° (linear pair)
\(\angle\)BOD = 180°- \(\angle\)AOD
\(\angle\)BOD = 180°- 75° = 105°
\(\angle\)CED =90° (angle in semi-circle)
\(\angle\)CDE =90° - \(\angle\)OCE \(\Rightarrow\) 90° - 40° = 50°
\(\angle\)OBE = \(\angle\)OBD
\(\angle\)OBD = 180°- (105° + 50°)
(In \(\Delta\)DBO, Angle sum property of \(\Delta\))
\(\angle\)OBE = \(\angle\)OBD = 25°

100°

8 cm

Given ED II BC
AB =AC
\(\Rightarrow\) \(\angle\)2 = \(\angle\)3 ...(i)
(Opp. Ls to opp. side are always equal)
In an isosceles II
\(\angle\)1 + \(\angle\)2 = 180° [Interior \(\angle\)s)
\(\therefore\) \(\angle\)3 + \(\angle\)4 = 180⁰
\(\therefore\) \(\angle\)1 + \(\angle\)3 = 180⁰
\(\angle\)2 + \(\angle\)4 = 180° 
but these are opp angles of a quad.
\(\therefore\) BCDE is a cyclic quad.

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.

Take a point S in the major arc. Join PS and RS.

∵ PQRS is a cyclic quadrilateral.
∴ \(\angle PQR+\angle PSR=180°\)
The sum of either pair- of opposite angles of a cyclic quadrilateral is 180°
\(\Rightarrow \ 100°+\angle PSR=180°\)
 \(\angle PSR=180°-100°\)
\(\Rightarrow \)  \(PSR=80°\)                  ....(1)
Now, ​​​​​​\(\angle POR=2\angle PSR\)
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
 \(2\times 80°=160°\)          ...(2) Using (1)
In \(\Delta OPR\),
∵   OP = OR        | Radii of a circle
∴ \(\angle OPR=\angle OPR\)           ....(3)
Angles opposite to equal sides of a triangle are equal
In \(\Delta OPR\),
\(\angle OPR+\angle ORP+\angle POR=180°\)
Sum of all the angles of a triangle is 180°
⇒ \(\angle OPR+\angle OPR+160°=180°\)  | Using (2) and (1)
⇒ ​​​​​\(2\angle OPR+160°=180°\)
⇒ ​​​​​ \(2\angle OPR=180°-160°=20°\)
⇒​​​​​ \(\angle OPR=10°\).

Given: A circle with centre O. Its two equal chords AB and CD intersect at E.
To prove: AE = DE and CE = BE.
Construction: Draw OM 丄 AB and ON 丄 CD join OE.
Proof: In \(\Delta OME\) and \(\Delta ONE\) ,
OM = ON
| ∵ Equal chords of a circle are equidistant from the centre
OE = OE                   I Common

∴\(\Delta OME\cong \Delta ONE\)  | RHS Rule
∴ ME = NE I CPCT
⇒AM + ME = DN + NE
∵ \(AB=CD\Rightarrow \frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } CD\Rightarrow AM=DN\)
⇒  AE = DE
⇒ AB - AE = CD - DE  | ∵ AB=CD
⇒  B E= CE.  | Given

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

∵ \(\Delta OAB\) is equilateral
∴ \(\angle AOB=60°\)
∴ \(\angle APB=\frac { 1 }{ 2 } \angle AOB=30°\)

See a circle

(i) Let us assume that A, Band C are the position of Priyanka, Sania and David respectively on the boundary of circular park with centre O.
Draw AD\(\bot\) BC
Since the centre of the circle coincides with the centroid of the equilateral \(\triangle\) ABC
\(\therefore\) Radius of circumscribed circle = \(\frac{2}{3}\) AD
\(\Rightarrow 20=\frac { 2 }{ 3 } AD\)
\(\Rightarrow AD=20\times \frac { 2 }{ 3 } \)
\(\Rightarrow AD=30m\)
Now, AD\(\bot\)BC, and let AB=BC=CA=x
\(\Rightarrow BD=CD=\frac { 1 }{ 2 } BC=\frac { x }{ 2 } \)

In rt. \(\triangle\)BDA, D=90⁰
By Pythagoras Theorem, we have
AB²=BD²+AD²
\(\Rightarrow { x }^{ 2 }={ \left( \frac { x }{ 2 } \right) }^{ 2 }+{ (30) }^{ 2 }\)
\(\Rightarrow{ x }^{ 2 }-{ \frac { { x }^{ 2 } }{ 4 } }=90\)
\(\Rightarrow \frac { 3 }{ 4 } { x }^{ 2 }=90\)
\(\Rightarrow { x }^{ 2 }=900\times \frac { 4 }{ 3 } \)
\(\Rightarrow { x }^{ 2 }=1200\)
\(\therefore x=\sqrt { 1200 } =20\sqrt { 3 } \)
Hence, the distance between each of them is \(20\sqrt { 3 } .\)
(ii) Properties of the circle, equilateral triangle and Pythagoras theorem.
(iii) Live and let live.