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Exercise 10.6

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

Sol. Consider triangles APB and AQB,

AB is common

∴ ΔAPB ≅  ΔAQB                                  [SSS]

∴   ∠APB = ∠AQB.                                [CPCT]

2. Two chords AB and CD of lengths 5 cm and 11 cm 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 6 cm, find the radius of the circle.

Sol.  Let OL is drawn perpendicular to AB and LO is produced to M meeting CD at M, then LM ⊥ CD.
[ AB ǁ‖ CD]

∴   ∠ALM = 90° ; ∠OMC = 90°

$AL = {1 \over 2}AB = {5 \over 2}cm;$

$CM = {1 \over 2}CD = {{11} \over 2}cm.$

Let OL = x, OA = r. Then OM = 6 – x.

Now, in right-angled triangle OLA.

$O{A^2} = A{L^2} + O{L^2} = {\left( {{5 \over 2}} \right)^2} + {X^2}$

$\Rightarrow {r^2} = {{25} \over 4} + {x^2}$                                 ……(i)

In right-angled triangle OML,

$O{C^2} = C{M^2} + O{M^2} = {\left( {{{11} \over 2}} \right)^2} + {(6 - x)^2}$

$\Rightarrow {r^2} = {{121} \over 4} + 36 - 12x + {x^2}$                   ……(ii)

From (i) and (ii), we get

${{25} \over 4} + {x^2} = {{121} \over 4} + 36 - 12x + {x^2}$

$\Rightarrow 12x = {{121} \over 4} + 36 - {{25} \over 4} = 6 \Rightarrow x = 5$

Substituting x = 5 in (i), we get

${r^2} = {{25} \over 4} + {(5)^2} = {{25} \over 4} + 25 = {{125} \over 4}$

$\Rightarrow r = {{5\sqrt 5 } \over 2}cm$

Hence, radius of the circle is ${{5\sqrt 5 } \over 2}cm$ .

3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Sol. Let AB = 6 cm and CD = 8 cm are two parallel chords of a circle with centre at O.
We draw OM perpendicular to AB meeting AB at M and CD at L. Then OL is also perpendicular to CD.

$AM = {1 \over 2}AB = {1 \over 2} \times 6 = 3$

and $CL = {1 \over 2}CD = {1 \over 2} \times 8 = 4cm.$

Let OL = x.

In right-angled triangle OMA,

OA2 = AM² + OM² = (3)² + (4)² = 25    ⇒  OA = 5 cm.

As OC = OA  ∴ OC = 5 cm.

In right-angled triangle OLC,

OC² = CL² + OL²  ∴ (5)² = (4)² + x²

⇒  x² = 25 – 16 = 9    ∴    x = 3

Therefore, distance of other chord is 3 cm.

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 [ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Sol. Construction: Join E and A.
Proof: ∠AOC = 2∠AEC                 …(i)

[Angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle.]

∠DOE = 2∠DAE                                     …(ii)                              [Reason same as above]

Also ∠AEC = ∠ABC + ∠BAE

[Exterior angle of a triangle is equal to sum of interior opposite angles]

⇒  ∠AEC = ∠ABC + ∠DAE

$\Rightarrow {1 \over 2}\angle AOC = \angle ABC + {1 \over 2}\angle DOE$

⇒  $\angle ABC = {1 \over 2}(\angle AOC - \angle DOE)$

5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Sol. We know that diagonals of a rhombus intersect each other at right angles.

∴  ∠AOB = 90°. Also we know that angle in a semicircle is 90°. If a circle is drawn with AB as diameter, then O lies on the circle.

Hence, if a circle is drawn with any sides of the rhombus as diameter, then it passes through the point of intersection of diagonals.

6. ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary) at E. Prove that AE = AD.

Sol. ABCD is a parallelogram.
∴  ∠ABC =  ∠CDA …(i)                                                        [Opposite angles of a parallelogram]

Also, ABCE is a cyclic quadrilateral.

∴  ∠ABC + ∠AEC = 180°                                                            …(ii)

[Sum of opposite angles of a cyclic quadrilateral is 180°]

In Fig. I,

⇒     ∠ABC + ∠ADE = 180° .                                                  ..(iii) [From (i)]

From (ii) and (iii), we get

[Sides opposite to equal angles of a triangle are equal.]

In Fig. II,
∠AEC + ∠AED = 180° …(iv)                                                [Linear pair]

From (i) and (ii), we get

∠CDA + ∠AEC = 180°    ⇒    ∠EDA + ∠AEC = 180°                                      …(v)

From (iv) and (v), we get

∠AED = ∠EDA

[Sides opposite to equal angles of a triangle are equal]

7. AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.

Sol. (i) Consider triangles AOD and BOC.

AO = OC [Given]

BO = OD [Given]

∠AOD = ∠BOC                          [Vertically opposite angles]

⇒   ΔAOD   ≅   ΔCOB                            [SAS]

⇒  ∠DAO = ∠BCO

[Angles in the same segment of a circle]
From (i) and (ii), we have

⇒  OA = OD

Hence, OA = OB = OC = OD.

⇒   0 is equidistant from A, B, C, D.

⇒   0 is centre of the circle.

Hence, AC and BD are diameters.

(ii) As AC and BD are diameters. Therefore, ∠BAD,∠ABC, ∠BCD, ∠CDA are angles in a semicircle.
Therefore, each of them is 90°.
Hence, ABCD is a rectangle.

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 $90^\circ - {1 \over 2}\angle A,90^\circ - {1 \over 2}\angle B$  and

[Angles in the same segment of a circle]
∠ADF = ∠ACF                                                …(ii) [Reason same as above]

⇒  ∠EDF = ∠ABE + ∠ACF

…(iii) [From (i), (ii)]

As BE and CF are bisectors of ∠ABC and ∠ACB.

$\angle ABE = {1 \over 2}\angle ABC$  and $\angle ACF = {1 \over 2}\angle ACB$   ….(iv)

From (iii) and (iv), we get

$\angle EDF = {1 \over 2}\angle ABC + {1 \over 2}\angle ACB = {1 \over 2}(\angle ABC + \angle ACB)$                                           …..(v)

In triangle ABC,

∠ABC + ∠ACB + ∠BAC   =180°

⇒ ∠ABC + ∠ACB  = 180° – ∠BAC.

Substituting in (v), we get

$\angle EDF = {1 \over 2}(180^\circ - \angle BAC) = 90^\circ - {1 \over 2}\angle BAC = 90^\circ - {1 \over 2}\angle A$

Similarly, we can show that
$\angle DEF = 90^\circ - {1 \over 2}\angle B;\angle DFE = 90^\circ - {1 \over 2}\angle C$

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 BP = BQ.

Sol. As AB is common chord.

∴   arc AXB = arc AYB
⇒  ∠AQB = ∠APB
[Equal areas of congruent circles subtend equal angles on the remaining part of the circles.]
Now in triangle BPQ,
∠AQB = ∠APB                                                [Proved above]
⇒  ∠PQB = ∠QPB
⇒  PB = QB.

[Sides opposite to equal angles are equal.]

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

Sol. Let 0 be the centre of the circumcircle of triangle ABC.
∴   ∠BOC = 2∠BAC
[Angle subtended by an arc at the centre is doubled the angle subtended by it at remaining part of the circle.]

Consider ΔOLB and ΔOLC.

OC is common.

BL = LC                                                                  [OL is perpendicular bisector]

∴   ΔOLB  ≅  ΔOLC                                               [SSS]

∴  ∠BOL = ∠COL = ∠CAB                                       …(i)

Let perpendicular bisector of side BC and angle bisector ∠A meet a the point P.
From equation (i), we get

∠COL = ∠CAB

⇒   ∠COP = 2∠CAP                                                            [∠CAP = ∠BAP]

This proves that points C, A and P are concyclic points as ∠COP is subtended by arc CP at the centre O and ∠CAP is subtended by it at the remaining part of the circle.
Again, from equation (i), we get

∠BOL = ∠BAC

i.e., ∠BOP = 2∠BAP

This, also proves that points B, A and P are concyclic. Hence, P lies on the circumcircle of the ΔABC.