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**Question 1. In the figure, OD is perpendicular to chord AB of a circle whose centre is O. If BC is a diameter, prove that CA = 2OD. Solution:** We have a circle whose centre is O. BC is a diameter and AB a chord such that OD ⊥ AB. Let us join AC.

∵ The perpendicular from centre of a circle to a chord bisects the chord

∴ D is the mid-point of AB

∵ O is the mid-point of the diameter BC

∴ OD is the line segment joining the mid-points of two sides of ΔABC.

∴ OD is half of the third side of ΔABC.

i.e. OD =(1/2)AC

or 2OD = AC

** Question 2. l is a line intersecting two concentric circles having common centre O, at A, B, C and D. Prove that AB = CD. Solution: **We have two concentric circles with common centre O.

Line l intersects these circles at A, B, C and D.

Let us draw OP ⊥ ℓ For the inner circle

∵ OP ⊥ BC [construction]

∴ P is the mid-point of BC [∵ Perpendicular from the centre to a chord bisects the chord]

∴ PB = PC …(1)

For the outer circle.

∵ OP ⊥ AD

∴ PA = PD ..(2)

Subtracting (1) from (2), we have PA - PB = PD - PC

⇒ AB = CD

**Question 3. AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED.**

**Solution: **We have a circle with centre O. Chord AB = chord CD and on production, AB and CD meet at E.

Let us join OE and draw OP ⊥ AB and OQ ⊥ CD.

∵ In a circle, equal chords are equidistant from the centre.

∴ OP = OQ. [∵ AB = CD]

Now, in right ΔOPE and right ΔOQE, we have OP = OQ [Proved],

OE = OE [Common]

∵ Using RHS criterion, ΔOPE ≌ ΔOQE

⇒ P E = QE [c.p.c.t.] …(1)

But AB = CD [Given]

⇒ (1/2) AB = (1/2)CD or PB = QD [∵ OP ⊥ AB and OQ ⊥ CD] …(2)

Subtracting (2) from (1), we have PE ∠ PB = QE ∠ QD

⇒ EB = ED

**Question 4. If O be the centre of the circle, find the value of ‘x’ in each of the following figures.**

**Solution: **(i) ∵ OA = OB [Radii of the same circle]

∴ ∠A = ∠B [Angles opposite to equal side in a triangle are equal]

In ΔABC, ∠A + ∠B + ∠O = 180º

∴ x + x + 70º = 180º [∵ ∠O = 70º (given) and ∠A = ∠B]

⇒ 2x + 70º = 180º

⇒ 2x = 180º - 70º = 110º

⇒ x= (110^{0}/2)= 55º

Thus, x = 55°

(ii) In ΔAOC, ∠A + ∠ACO + ∠AOC = 180º

⇒ 40º + ∠ACO + 90º = 180º

⇒ ∠ACO = 180º - 40º - 90º = 50º

∵ AB is a diameter.

∴ ∠ACB = 90º [Angle in a semicircle]

∴ 50º + x = 90º

⇒ xº = 90º - 50º = 40º

Thus, x = 40º

(iii) ∵ ∠AOC + ∠COB = 180º [Linear pairs]

∴ 120º + ∠COB = 180º

⇒ ∠COB = 180º - 120º = 60º

∵ The arc CB is subtending ∠COB at the centre and ∠CDB at the remaining part.

∴ ∠CDB = (1/2)∠COB

⇒ x= (1/2)(60º) = 30º

⇒ x = 30º

(iv) In ΔAOC,

∵ AO = OC [Radii of the same circle]

∴ ∠OAC = ∠OCA [Angles opposite to equal sides are equal]

⇒ ∠OAC = 50º

∴ Exterior ∠COB = 50º + 50º = 100º.

Now, the arc BC is subtending ∠BOC at the centre and ∠BDC at the remaining part of the circle.

∴ ∠BDC = (1/2)∠BOC

⇒ x= (1/2)(100º) = 50º

Thus, x = 50º

(v) In ΔOAC, OA = OC [Radii of the same circle]

∴ ∠AOC = ∠ACO [∵ Angles opposite to equal sides are equal]

Now, ∠AOC + ∠ACO + ∠OAC = 180º

⇒ ∠AOC + ∠ACO + 50º = 180º

⇒ ∠AOC + ∠ACO = 180º - 50º = 130º

⇒ ∠AOC = ∠ACO = (130^{0}/2) = 65º

Now, ∠AOB + ∠AOC = 180º [Linear pairs]

∴ ∠AOB + 65º = 180º

⇒ ∠AOB = 180º - 65º = 125º

∵ The arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.

∴ ∠ADB = (1/2) ∠AOB

⇒ x= (1/2)(125º) = 62 (1/2)

∴ x= 62(1/2)

(vi) ∵ ∠BDC = ∠BAC [Angles in the same segment]

∴ ∠BDC = 40º

Now, in ΔBDC, we have ∠BDC + ∠CBD + ∠BCD = 180º

∴ 40º + 80º + x = 180º

⇒ 120º + x = 180º

⇒ x = 180º - 120º = 60º

Thus, x = 60º

**Question 5. In the adjoining figure, O is the centre of the circle. Prove that ∠ XOZ = 2(∠ XZY + ∠ YXZ). Solution:** Let us join OY.

∵ The arc XY subtends ∠XOY at the centre and ∠XZY at a point Z on the remaining part of the circle.

∴ ∠XOY = 2∠XZY …(1)

Similarly, ∠YOZ = 2∠YXZ …(2)

Adding (1) and (2), we have ∠XOY + ∠YOZ = 2∠XZY + 2∠YXZ

⇒ ∠XOZ = 2[∠XZY + ∠YXZ]

**Question 6. Show that the sum of the opposite angles of a cyclic quadrilateral is 180º. Solution:** We have a cyclic quadrilateral. Let us join AC and BD. Since, angles in the same segment are equal.

∴ ∠ACB = ∠ADB …(1)

and ∠BAC = ∠BDC …(2)

Adding (1) and (2), we have

∠ACB + ∠BAC = ∠ADB + ∠BDC

⇒ ∠ACB + ∠BAC = ∠ADC

Adding ∠ABC to both sides, we have ∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC

But, ∠ACB + ∠BAC + ∠ABC = 180º [Sum of the angles of ΔABC = 180º]

∴ ∠ADC + ∠ABC = 180º

⇒ ∠B + ∠D = 180º

Since, ∠A + ∠B + ∠C + ∠D = 360º

⇒ ∠A + ∠C = 360º ∠ 180º = 180º

**Question 7. Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Solution: **We have a cyclic quadrilateral ABCD in which the bisectors ∠A, ∠B, ∠C and ∠D for a quadrilateral PQRS.

From ΔABP, we have ∠PAB + ∠PBA + ∠P = 180º [Sum of the three angles of ΔABP]

⇒ (1/2) ∠A + (1/2)∠B + ∠P = 180º …(1)

From ΔCDR, we have

∠RCD + ∠RDC + ∠R = 180º [Sum of the three angles of ΔCDR.]

⇒ (1/2)∠C +(1/2)∠D + ∠R = 180° …(2)

Adding (1) and (2), we have (1/2)∠A + (1/2)∠B + (1/2)∠C +(1/2)∠D + ∠P + ∠R = 360º

⇒ (1/2) (∠A + ∠B + ∠C + ∠D) + ∠P + ∠R = 360º

⇒ (1/2)(360º) + ∠P + ∠R = 360º [∵ ∠A + ∠B + ∠C + ∠D = 360°]

⇒ ∠P + ∠R = 360º -(1/2) (360º) = 180º

Similarly, ∠Q + ∠S = 180º

Thus, the pairs of opposite angles of quadrilateral PQRS are supplementary.

Hence, PQRS is cyclic.

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