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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 âŠ¥ AB. Let us join AC.

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

âˆ´ D is the mid-point of AB

âˆµ O is the mid-point of the diameter BC

âˆ´ OD is the line segment joining the mid-points of two sides of Î”ABC.

âˆ´ OD is half of the third side of Î”ABC.

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 âŠ¥ â„“ For the inner circle

âˆµ OP âŠ¥ BC [construction]

âˆ´ P is the mid-point of BC [âˆµ Perpendicular from the centre to a chord bisects the chord]

âˆ´ PB = PC â€¦(1)

For the outer circle.

âˆµ OP âŠ¥ AD

âˆ´ PA = PD ..(2)

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

â‡’ 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 âŠ¥ AB and OQ âŠ¥ CD.

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

âˆ´ OP = OQ. [âˆµ AB = CD]

Now, in right Î”OPE and right Î”OQE, we have OP = OQ [Proved],

OE = OE [Common]

âˆµ Using RHS criterion, Î”OPE â‰Œ Î”OQE

â‡’ P E = QE [c.p.c.t.] â€¦(1)

But AB = CD [Given]

â‡’ (1/2) AB = (1/2)CD or PB = QD [âˆµ OP âŠ¥ AB and OQ âŠ¥ CD] â€¦(2)

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

â‡’ EB = ED

**Question 4. If O be the centre of the circle, find the value of â€˜xâ€™ in each of the following figures.**

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

âˆ´ âˆ A = âˆ B [Angles opposite to equal side in a triangle are equal]

In Î”ABC, âˆ A + âˆ B + âˆ O = 180Âº

âˆ´ x + x + 70Âº = 180Âº [âˆµ âˆ O = 70Âº (given) and âˆ A = âˆ B]

â‡’ 2x + 70Âº = 180Âº

â‡’ 2x = 180Âº - 70Âº = 110Âº

â‡’ x= (110^{0}/2)= 55Âº

Thus, x = 55Â°

(ii) In Î”AOC, âˆ A + âˆ ACO + âˆ AOC = 180Âº

â‡’ 40Âº + âˆ ACO + 90Âº = 180Âº

â‡’ âˆ ACO = 180Âº - 40Âº - 90Âº = 50Âº

âˆµ AB is a diameter.

âˆ´ âˆ ACB = 90Âº [Angle in a semicircle]

âˆ´ 50Âº + x = 90Âº

â‡’ xÂº = 90Âº - 50Âº = 40Âº

Thus, x = 40Âº

(iii) âˆµ âˆ AOC + âˆ COB = 180Âº [Linear pairs]

âˆ´ 120Âº + âˆ COB = 180Âº

â‡’ âˆ COB = 180Âº - 120Âº = 60Âº

âˆµ The arc CB is subtending âˆ COB at the centre and âˆ CDB at the remaining part.

âˆ´ âˆ CDB = (1/2)âˆ COB

â‡’ x= (1/2)(60Âº) = 30Âº

â‡’ x = 30Âº

(iv) In Î”AOC,

âˆµ AO = OC [Radii of the same circle]

âˆ´ âˆ OAC = âˆ OCA [Angles opposite to equal sides are equal]

â‡’ âˆ OAC = 50Âº

âˆ´ Exterior âˆ COB = 50Âº + 50Âº = 100Âº.

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

âˆ´ âˆ BDC = (1/2)âˆ BOC

â‡’ x= (1/2)(100Âº) = 50Âº

Thus, x = 50Âº

(v) In Î”OAC, OA = OC [Radii of the same circle]

âˆ´ âˆ AOC = âˆ ACO [âˆµ Angles opposite to equal sides are equal]

Now, âˆ AOC + âˆ ACO + âˆ OAC = 180Âº

â‡’ âˆ AOC + âˆ ACO + 50Âº = 180Âº

â‡’ âˆ AOC + âˆ ACO = 180Âº - 50Âº = 130Âº

â‡’ âˆ AOC = âˆ ACO = (130^{0}/2) = 65Âº

Now, âˆ AOB + âˆ AOC = 180Âº [Linear pairs]

âˆ´ âˆ AOB + 65Âº = 180Âº

â‡’ âˆ AOB = 180Âº - 65Âº = 125Âº

âˆµ The arc AB subtends âˆ AOB at the centre and âˆ ADB at the remaining part of the circle.

âˆ´ âˆ ADB = (1/2) âˆ AOB

â‡’ x= (1/2)(125Âº) = 62 (1/2)

âˆ´ x= 62(1/2)

(vi) âˆµ âˆ BDC = âˆ BAC [Angles in the same segment]

âˆ´ âˆ BDC = 40Âº

Now, in Î”BDC, we have âˆ BDC + âˆ CBD + âˆ BCD = 180Âº

âˆ´ 40Âº + 80Âº + x = 180Âº

â‡’ 120Âº + x = 180Âº

â‡’ x = 180Âº - 120Âº = 60Âº

Thus, x = 60Âº

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

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

âˆ´ âˆ XOY = 2âˆ XZY â€¦(1)

Similarly, âˆ YOZ = 2âˆ YXZ â€¦(2)

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

â‡’ âˆ XOZ = 2[âˆ XZY + âˆ YXZ]

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

âˆ´ âˆ ACB = âˆ ADB â€¦(1)

and âˆ BAC = âˆ BDC â€¦(2)

Adding (1) and (2), we have

âˆ ACB + âˆ BAC = âˆ ADB + âˆ BDC

â‡’ âˆ ACB + âˆ BAC = âˆ ADC

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

But, âˆ ACB + âˆ BAC + âˆ ABC = 180Âº [Sum of the angles of Î”ABC = 180Âº]

âˆ´ âˆ ADC + âˆ ABC = 180Âº

â‡’ âˆ B + âˆ D = 180Âº

Since, âˆ A + âˆ B + âˆ C + âˆ D = 360Âº

â‡’ âˆ A + âˆ C = 360Âº âˆ 180Âº = 180Âº

**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 âˆ A, âˆ B, âˆ C and âˆ D for a quadrilateral PQRS.

From Î”ABP, we have âˆ PAB + âˆ PBA + âˆ P = 180Âº [Sum of the three angles of Î”ABP]

â‡’ (1/2) âˆ A + (1/2)âˆ B + âˆ P = 180Âº â€¦(1)

From Î”CDR, we have

âˆ RCD + âˆ RDC + âˆ R = 180Âº [Sum of the three angles of Î”CDR.]

â‡’ (1/2)âˆ C +(1/2)âˆ D + âˆ R = 180Â° â€¦(2)

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

â‡’ (1/2) (âˆ A + âˆ B + âˆ C + âˆ D) + âˆ P + âˆ R = 360Âº

â‡’ (1/2)(360Âº) + âˆ P + âˆ R = 360Âº [âˆµ âˆ A + âˆ B + âˆ C + âˆ D = 360Â°]

â‡’ âˆ P + âˆ R = 360Âº -(1/2) (360Âº) = 180Âº

Similarly, âˆ Q + âˆ S = 180Âº

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

Hence, PQRS is cyclic.

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