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#### Q : 10     In any triangle ABC, if the angle bisector of   and perpendicular bisector of BC                intersect, prove that they intersect on the circumcircle of the triangle ABC.

Given :In any triangle ABC, if the angle bisector of   and perpendicular bisector of BC  intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let ABD = 1 ,  ADC = 2 , DCB =3 , CBD = 4

1 and 3 lies in same segment.So,

1 = 3    ..........................1(angles in same segment)

similarly, 2 = 4  ......................2

also,       1=2 ..............3(given)

From 1,2,3 , we get

3 = 4

Hence,  BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

#### Q: 9        Two congruent circles intersect each other at points A and B. Through A any line                 segment PAQ is drawn so that P, Q lie on the two circles. Prove that .

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove :  BP = BQ

Proof :

AB is a common chord in both congruent circles.

In

(Sides opposite to equal of the triangle are equal )

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#### Q : 8       Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and                 F respectively. Prove that the angles of the triangle DEF are    ,   and

Given :   Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove :  the angles of the triangle DEF are    ,     and

Proof :

1 and 3 are angles in same segment.therefore,

1 = 3 ................1(angles in same segment are equal )

and    2 = 4 ..................2

1+2=3+4

,

and

Similarly,       and

#### Q : 6     ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if              necessary) at E. Prove that .

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove :   AE = AD

Proof :

ADC = 3  , ABC = 4, ADE = 1  and AED = 2

.................1(linear pair)

....................2(sum of opposite angles of cyclic quadrilateral)

3 = 4      (oppsoite angles of parallelogram )

From 1 and 2,

3+1 = 2 +

From 3,    1 =

From 4,    AQB,     1 = 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

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#### Q: 4        Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

To prove :

Construction: Join AC and DE.

Proof :

Let ADC = x , DOE = y and  AOD = z

So,  EOC = z (each chord subtends equal angle at centre)

AOC + DOE +AOD + EOC =

.........................................1

OA = OD  (Radii of the circle)

OAD = ODA    (angles opposite to equal sides )

OAD + ODA + AOD =

.............................................................2

Similarly,

.............................................................3

..............................................................4

ODB is exterior of triangle OAD . So,

(from  2)

.................................................................5

similarly,

OBE is exterior of triangle OCE . So,

OBE = OCE + OEC

(from  3)

.................................................................6

From 4,5,6 ;we get

BDE = BED = OEB - OED

..................................................7

In BDE ,

DBE + BDE + BED =

...................................................8

Here,  from equation 1,

...................................9

From 8 and 9,we have

#### Q : 3  The lengths of two parallel chords of a circle are and . If the smaller chord is at distance from the centre, what is the distance of the other chord from the centre?

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw

Proof :

Proof: CD is a chord of circle  and

Thus, CM = MD = 3 cm    (perpendicular from centre bisects chord)

and      AN = NB  = 4 cm

Let MN be x.

So, ON = 4 - x              (MN = 4 cm )

In  OCM , using Pythagoras,

.............................1

and

In  OAN , using Pythagoras,

.............................2

From 1 and 2,

So, x=1  (since  )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

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#### Q: 2     Two chords AB and CD of lengths and respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is , find the radius of the circle.

Given : AB = 5 cm, CD = 11 cm and AB || CD.

Construction: Draw

Proof :

Proof: CD is a chord of circle  and

Thus, CM = MD = 5.5 cm    (perpendicular from centre bisects chord)

and      AN = NB  = 2.5 cm

Let OM be x.

So, ON = 6 - x              (MN = 6 cm )

In  OCM , using pythagoras,

.............................1

and

In  OAN , using pythagoras,

.............................2

From 1 and 2,

From 2, we get

OA = OC

Thus, the radius of the circle is

#### Q : 7    AC and BD are chords of a circle which bisect each other. Prove that           (ii) ABCD is a rectangle.

Given :  AC and BD are chords of a circle which bisect each other.

To prove : ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

ABCD is a parallelogram . (proved in (i))

(proved in (i))

A parallelogram with one angle , is a rectangle )

Thus, ABCD is rectangle.

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#### Q: 7    AC and BD are chords of a circle which bisect each other. Prove that           (i) AC and BD are diameters

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

In  ABD and CDO,

AO = OC     (Given )

AOB = COD  (Vertically opposite angles )

BO = DO        (Given )

So,   ABD CDO     (By SAS)

BAO = DCO    (CPCT)

BAO and DCO are alternate angle and are equal .

So,     AB || DC ..............1

From 1 and 2,

......................3(sum of opposite angles)

A = C    ................................4(Opposite angles of the parallelogram )

From 3 and 4,

BD is a diameter of the circle.

Similarly, AC is a diameter.

#### Q: 1    Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : PAQ = PBQ

Proof : In APQ and BPQ,

PA = PB         (radii of same circle)

PQ = PQ        (Common)

QA = QB       (radii of same circle)

So,           APQ  BPQ       (By SSS)

PAQ = PBQ     (CPCT)