PROBLEM – 1. ABCD is quadrilateral inscribed in a circle with centre O. If the diagonals AC and BD intersect at the point E,prove that ∠AOB+∠COD=2∠AER.

** General Enunciation**: ABCD is quadrilateral inscribed in a circle with centre O. If the diagonals AC and BD intersect at the point E, we have to prove that ∠AOB + ∠COD=2∠AEB.

**Particular Enunciation**: Given that two chords AC and BD of a circle with centre O, intersect in the interior point E of the circle .Let us join (O, A), (O,B),(O,C) and (O,D). We have to prove that

(∠AOB+ <COD) =2∠AEB.

** Proof**: ∠AOB and ∠ADB are two angles at the centre and at the circumference respectively on the same are AB.

(∠AOB=2∠ADB) ———————– (1)

Again, ∠COD and ∠CAD are two angles at the centre and at the circumference respectively on the same are CD.

∠COD=2∠CAD ————————– (2)

Adding equations (1) and (2) we get

∠AOB+ ∠COD=2(∠ADB+∠CAD) … …(iii)

But in **∆**AED, external ∠AEB=∠EAD+∠EDA

∠AEB=∠CAD+∠ADB

Putting this value in equation (iii) we get, (∠AOB +∠COD)=2∠AEB __(proved).__

PROBLEM – 2. **Two chords AB and CD of the circle ABCD intersect at the point E. show that ∆AED and ∆BEC are equiangular.**

**General enunciation**: Two chords AB and CD of the circle ABCD intersect at the point E. we have two show that **∆**AED and **∆** BEC are equiangular.

**Particular enunciation**: Suppose, O is the centre of the circle ABCD. Two chords AB and CD intersect at E. (A,D)and (C,B) are joined. Let us show that **∆** AED and **∆** BEC are equiangular.

**Proof**: ∠DAB and ∠BCD are one some are BD.

∴∠ DAB=∠BCD

**Similarly**,∠ADC and∠ABC are on the some are AC.

∴ ∠ADC =∠ABC Now in **∆** AED and **∆** BEC we get

∠DAE= ∠BCE **[they line on the some arc BC]**

∠ADE=∠CBE **[they line on the some arc AC]**

And ∠ADE=∠BEC **[Vertically opposite angles]** __(Showed)__

** **PROBLEM – **3. In the circle ABCD with the centre O,**∠**ADB+**∠**BDC=1 right angle. Prove that A ,O and C lie in the some straight line**.

**General Enunciation:** In the circle ABCD with the centre O, ∠ADB+∠BDC=1right angle.

We have to prove that A, O and C line in the some straight line.

**Particular Enunciation**: Given that in the circle ABCD with the centre O, ∠ADB+ ∠BDC =1 right angle. Let us prove that A, O and C line in some straight line.

**Construction: **(A, O), (C,O) and (B,O) are joined**.**

**Proof**: ∠AOB and ∠ADB are two angles at the centre and at the circumference respectively on the some are AB.

∴ ∠AOB =2∠ADB ————————- (1)

**Similarl**y, ∠BOC and ∠BDC are two angles at the centre and at the circumference respectively on the same BC.

∴ ∠BOC = 2∠BDC ————————– (2)

By adding equations (1) and (2) we get,

∠AOB +∠BOC= 2 ∠ADB = ∠BDC )

∴∠AOB + ∠BOC =2 (right angle) [∠ADB+ ∠BDC= 1 right angle]

=2 (right angles, i.e. 1 straight angle.

Hence, A, O and C are one the same line __(Proved____)__

PROBLEM – 4**. In interior of a circle, two chords AB and CD interest at point E. Prove that sum of the angles subtended by the arcs AC and BD at the centre is twice **∠**AEC **

**General Enunciation**: Interior of a circle, two chords AB and CD intersect at a point E. We have to prove that the sum of the angles subtended by the arcs AC and BD at the centre is twice ∠AEC

**Particular Enunciation**: Given that two chords AB and CD of the circle ABCD with centre O intersect in the interior point E of the circle. Let us join (O, A), (O, B) (O, C) and (O,D).Two arcs AC and BD subtended two angles <AOC and <BOD respectively at the centre. Let us prove that (∠BOD+ ∠AOC) =2∠AEC

**Construction**: Let as join (B, C)

** Proof**: ∠AOC and ∠ABC are two angles at the centre and at the circumference respectively on the same arc AC.

∴∠AOC=2∠ABC —————————- (1)

Again, ∠BOD and ∠BCD are two angles at the centre and at the circumference respectively on the same arc BD

∴ ∠BOD=2∠BCD ————————– (2)

Adding equation (1) and (2) We get,

∠AOC+ <BOD=2(∠ABC+∠BCD) … … (3)

But in **∆**BCE, external ∠AEC=∠BCE+∠CBE

∴∠AEC = ∠BCD+∠ABC

Putting this value in equation (3)We get,

(∠AOC+∠BOD) =2∠AEC __(Proved)__

PROBLEM – **5.** **Show that the oblique sides of cyclic trapezium are equal**.

** General Enunciation**: We have show that the oblique sides of a cyclic trapezium are equal.

Particular Enunciation: Suppose, ABCD is the circle with the centre O. ABCD be the trapezium inscribed in that circle. In trapezium ABCD, AB ∥ CD and AD and BC are oblique. Let us show that AD=BC

**Construction**: Let us join (O, A),(O,B),(O,C),(O,D) and (B,D).

**Proof: **∠BOC and ∠BDC are two angles at the centre and at circumference respectively one some are BD.

∴∠BOC =2∠BDC ————————- (1)

Again, ∠AOD and ABD are two angles at the centre and at circumference respectively on the same are AD

∴∠AOD =2∠ABD ———————————(2)

But AB ∥ CD and BD is there scant.

∴ ∠ABD = ∠BDC

∴ ∠AOD= 2∠BDC ——————————-(3)

∴ From equations (1) and (3) we get ∠BOC =∠AOD

∴Arc BC= arc AD

∴ BC=AD

Hence BC=AD __(Showed)__

PROBLEM – **6. AB and AC are two chords of a circle; P and Q are the middle points of the two the minor arcs intersected by them. The chord PQ intersects the chords AB and AC at the points D and E respectively. Show that AD =AE. **

**General Enunciation**: ** **AB and AC are two chords of a circle; P and Q are the middle points of the two the minor arcs intersected by them. The chord PQ intersects the chords AB and AC at the points D and E respectively. we have to Show that

AD =AE**. **

**Particular Enunciation**: Suppose, in the circle; ABC with the centre O, AB and AC are two chords. P and Q are the middle points of the two minor arcs intersected by AB and AC respectively. The chord PQ intersects the chords AB and AC at the points d and E respectively. Let us show that AD = AE.

**Construction: **(A, P) and (P, C) are joined.

**Proof: **Since, P is the middle points of are AB,

∴ Are AP= are PB

∴ ∠ACP= ∠PAB ———————- (1) ** **

Again Q is the middle point of are AC

∴Arc AQ= arc QC

∴ ∠ACP+∠APQ …. …. (2)

By adding equations (1) and (2) we get,

∠ACP+∠CPQ=∠PAB ∠APQ ————————- (3)

But in **∆** PCE, external ∠ADQ=ECP+∠EPC

i,e., ∠AED= ∠ACP+∠CPQ —————————— (4)

Again in **∆** PAD,

external ∠ADQ =∠PAD=∠APD

Or, ∠ADE= ∠PAB+∠APQ

= ∠ACP+∠CPQ [From (3)]

=∠AED [from (4)]

∴ ∠ADE = ∠AED

Now in **∆**ADE since ∠ADE= AED

∴ AD= AE __(Showed)__