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

Let $A E$ be the angle bisector of $\angle A$.
To prove: ED is the perpendicular bisector of BC .
$\angle B A E=\angle C A E$-------- (1) [Since, $A E$ is the angle bisector of $\angle A]$
Now, $\angle E B C=\angle C A E$-------(2) [Angles subtended by the same arc EC]
Also, $\angle \mathrm{ECB}=\angle \mathrm{BAE}$------(3) [Angles subtended by the same arc BE]
But we know that, $\angle B A E=\angle C A E$ From equation (1)]
Hence, $\angle E B C=\angle E C B$ [From equations (2) and (3)]
Therefore, $\mathrm{BE}=\mathrm{EC}$ [Sides opposite to equal angles are equal]

The point E is equidistant from the points B and C .
Therefore, ED is the perpendicular bisector of BC.
The perpendicular bisector of BC and the angle bisector of $\angle A$ meet the circumcircle of $\triangle A B C$ at E.

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seema garhwal

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 $B P=B Q$.

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.

$\therefore \angle A P B=\angle A Q B$
$\operatorname{In} \triangle B P Q$,
$\angle A P B=\angle A Q B$
$\therefore B Q=B P$      (Sides opposite to equal of the triangle are equal)

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mansi

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Q 8. Bisectors of angles $\mathrm{A}, \mathrm{B}$ and C of a triangle ABC intersect its circumcircle at $\mathrm{D}, \mathrm{E}$ and F respectively.
Prove that the angles of the triangle DEF are $90^{\circ}-\frac{1}{2} C,90^{\circ}-\frac{1}{2} B$ and  $90^{\circ}-\frac{1}{2} A$

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 $90^{\circ}-\frac{1}{2} C,90^{\circ}-\frac{1}{2} B$ and  $90^{\circ}-\frac{1}{2} A$

Proof: 

$\angle 1$ and $\angle 3$ are angles in the same segment. Therefore,
$\angle 1=\angle 3$  (angles in same segment are equal)-------(1) 
and $\angle 2=\angle 4$-------2
Adding 1 and 2 , we have
$\angle 1+\angle 2=\angle 3+\angle 4$
$\Rightarrow \angle D=\frac{1}{2} \angle B+\frac{1}{2} \angle C$
$\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)$

$\Rightarrow \angle D=\frac{1}{2}\left(180^{\circ}+\angle C\right)$
and $\Rightarrow \angle D=\frac{1}{2}\left(180^{\circ}-\angle A\right)$
$\Rightarrow \angle D=90^{\circ}-\frac{1}{2} \angle A$
Similarly,
$\angle E=90^{\circ}-\frac{1}{2} \angle B$ and $\angle F=90^{\circ}-\frac{1}{2} \angle C$

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seema garhwal

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

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

To prove:   AE = AD 

Proof:

$\angle \mathrm{ADC}=\angle 3, \angle \mathrm{ABC}=\angle 4, \angle \mathrm{ADE}=\angle 1$ and $\angle \mathrm{AED}=\angle 2$
$\angle 3+\angle 1=180^{\circ}$  (linear pair)--------(1)
$\angle 2+\angle 4=180^{\circ}$(sum of opposite angles of cyclic quadrilateral)-----------(2)
$\angle 3=\angle 4$   (opposite angles of parallelogram )
From 1 and 2,
$\angle 3+\angle 1=\angle 2+\angle 4$
From 3, $\angle 1=\angle 2$
From 4, $\triangle A Q B$
$\angle 1=\angle 2$

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

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mansi

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

Given : $\mathrm{AD}=\mathrm{CE}$
To prove :
$\angle A B C=\frac{1}{2}(\angle A O C-\angle D O E)$
Construction: Join AC and DE.
Proof:

Let $\angle \mathrm{ADC}=\mathrm{x}, \angle \mathrm{DOE}=\mathrm{y}$ and $\angle \mathrm{AOD}=\mathrm{z}$
So, $\angle E O C=z$ (each chord subtends an equal angle at the centre)
$\angle \mathrm{AOC}+\angle \mathrm{DOE}+\angle \mathrm{AOD}+\angle \mathrm{EOC}=360^{\circ}$
$\Rightarrow x+y+z+z=360^{\circ}$
$\Rightarrow x+y+2 z=360^{\circ}$---------(1)
$\text { In } \triangle \mathrm{OAD}$,
$\mathrm{OA}=\mathrm{OD}$ (Radii of the circle)
$\angle \mathrm{OAD}=\angle \mathrm{ODA}$(angles opposite to equal sides)

$\angle \mathrm{OAD}+\angle \mathrm{ODA}+\angle \mathrm{AOD}=180^{\circ}$
$\Rightarrow 2 \angle O A D+z=180^{\circ}$
$\Rightarrow 2 \angle O A D=180^{\circ}-z$
$\Rightarrow \angle O A D=\frac{180^{\circ}-z}{2}$
$\Rightarrow \angle O A D=90^{\circ}-\frac{z}{2}$-------(2)

Similarly,

$\Rightarrow \angle O C E=90^{\circ}-\frac{x}{2}$------(3)
$\Rightarrow \angle O E D=90^{\circ}-\frac{y}{2}$------(4)
$\angle O D B$ is the exterior of triangle OAD. 
So, $\angle \mathrm{ODB}=\angle \mathrm{OAD}+\angle \mathrm{ODA}$
$\Rightarrow \angle O D B=90^{\circ}-\frac{z}{2}+z$   (from 2)
$\Rightarrow \angle O D B=90^{\circ}+\frac{z}{2}$-------(5)

Similarly,

$\angle O B E$ is the exterior of triangle OCE. 
So, $\angle \mathrm{OBE}=\angle \mathrm{OCE}+\angle \mathrm{OEC}$
$\Rightarrow \angle O E B=90^{\circ}-\frac{z}{2}+z$    (from 3)
$\Rightarrow \angle O E B=90^{\circ}+\frac{z}{2}$------(6)
From 4,5,6, we get
$\angle \mathrm{BDE}=\angle \mathrm{BED}=\angle \mathrm{OEB}-\angle \mathrm{OED}$
$\Rightarrow \angle B D E=\angle B E D=90^{\circ}+\frac{z}{2}-\left(90-\frac{y}{2}\right)=\frac{y+z}{2}$

\Rightarrow \angle BDE+\angle BED=y+z..................................................7

$\Rightarrow \angle B D E+\angle B E D=y+z$---------(7)
In $\triangle \mathrm{BDE}$
$\angle \mathrm{DBE}+\angle \mathrm{BDE}+\angle \mathrm{BED}=180^{\circ}$
$\Rightarrow \angle D B E+y+z=180^{\circ}$
$\Rightarrow \angle D B E=180^{\circ}-(y+z)$
$\Rightarrow \angle A B C=180^{\circ}-(y+z)$--------(8)
Here, from equation 1 ,
$\frac{x-y}{2}=\frac{360^{\circ}-y-2 x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360^{\circ}-2 y-2 x}{2}$
$\Rightarrow \frac{x-y}{2}=180^{\circ}-y-x$-------(9)
From 8 and 9, we have
$\angle A B C=\frac{x-y}{2}=\frac{1}{2}(\angle A O C-\angle D O E)$

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mansi

Q 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?

Given : $A B=8 \mathrm{~cm}, C D=6 \mathrm{~cm}, O M=4 \mathrm{~cm}$ and $A B \| C D$.
To find: Length of ON
Construction: Draw $O M \perp C D$ and $O N \perp A B$
Proof:

Proof: CD is a chord of a circle and $O M \perp C D$
Thus, $C M=M D=3 \mathrm{~cm} \quad$ (perpendicular from centre bisects chord) and $\mathrm{AN}=\mathrm{NB}=4 \mathrm{~cm}$
Let $M N$ be x .
So, ON = $4-\mathrm{x}$
$(\mathrm{MN}=4 \mathrm{~cm})$
In $\triangle$ OCM, using Pythagoras,
$O C^2=C M^2+O M^2$--------1

and

In $\triangle$ OAN, using Pythagoras,
$O A^2=A N^2+O N^2$
From 1 and 2,
$C M^2+O M^2=A N^2+O N^2$
$\Rightarrow 3^2+4^2=4^2+(4-x)^2$
$\Rightarrow 9+16=16+16+x^2-8 x$
$\Rightarrow 9=16+x^2-8 x$
$\Rightarrow x^2-8 x+7=0$

$\Rightarrow x^2-7 x-x+7=0$
$\Rightarrow x(x-7)-1(x-7)=0$
$\Rightarrow(x-1)(x-7)=0$
$\Rightarrow x=1,7$
So, $\mathrm{x}=1$ (since $x \neq 7>O M$ )
$\mathrm{ON}=4-\mathrm{x}=4-1=3 \mathrm{~cm}$

Hence, the second chord is 3 cm away from the centre.

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seema garhwal

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Q 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 $A B$ and $C D$ is 6 cm , find the radius of the circle.

Given: $A B=5 \mathrm{~cm}, C D=11 \mathrm{~cm}$ and $A B \| C D$.
To find Radius (OA).
Construction: Draw $O M \perp C D$ and $O N \perp A B$
Proof :

          

Proof: CD is a chord of the circle and $O M \perp C D$
Thus, $\mathrm{CM}=\mathrm{MD}=5.5 \mathrm{~cm}$ (perpendicular from centre bisects chord)
and $\mathrm{AN}=\mathrm{NB}=2.5 \mathrm{~cm}$
Let $O M$ be x .
So, $\mathrm{ON}=6-\mathrm{x} \quad(\mathrm{MN}=6 \mathrm{~cm})$
In $\triangle$ OCM, using Pythagoras,
$O C^2=C M^2+O M^2$-----------1

and

In $\triangle$ OAN, using Pythagoras,
$O A^2=A N^2+O N^2$------(2)
From 1 and 2,
$C M^2+O M^2=A N^2+O N^2$ (OC=OA= radii)
$5.5^2+x^2=2.5^2+(6-x)^2$
$\Rightarrow 30.25+x^2=6.25+36+x^2-12 x$
$\Rightarrow 30.25-42.25=-12 x$
$\Rightarrow-12=-12 x$

$\Rightarrow x=1$
From 2, we get
$O C^2=5.5^2+1^2=30.25+1=31.25$
$\Rightarrow O C=\frac{5}{2} \sqrt{5} \mathrm{~cm}$
$\mathrm{OA}=\mathrm{OC}$
Thus, the radius of the circle is $\frac{5}{2} \sqrt{5} \mathrm{~cm}$.

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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))

\angle A=90 \degree                    (proved in (i))

A parallelogram with one angle 90 \degreeis a rectangle )

Thus, ABCD is a rectangle.   

 

 

 

 

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Q 7. AC and BD are chords of a circle that 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 $\triangle \mathrm{ABD} \text { and } \triangle \mathrm{CDO}$,
$\mathrm{AO}=\mathrm{OC}$-----(Given)$
$\angle \mathrm{AOB}=\angle \mathrm{COD}$---- (Vertically opposite angles)
$\mathrm{BO}=\mathrm{DO}$----(Given)
So, $\triangle \mathrm{ABD} \cong \triangle \mathrm{CDO}$  (By SAS)
$\angle \mathrm{BAO}=\angle \mathrm{DCO} 
 ---(\mathrm{CPCT})$
$\angle \mathrm{BAO}$ and $\angle \mathrm{DCO}$ are alternate angles and are equal.
So, $A B \| D C$ ----------1
Also $A D \| B C$ ---------- 2

From 1 and 2,

$\angle A+\angle C=180^{\circ}$
$\angle \mathrm{A}=\angle \mathrm{C}$
From 3 and 4,
$\angle A+\angle A=180^{\circ}$
$\Rightarrow 2 \angle A=180^{\circ}$
$\Rightarrow \angle A=90^{\circ}$

BD is the diameter of the circle.

Similarly, AC is a diameter.

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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 : \anglePAQ = \anglePBQ 

Proof : In \triangleAPQ and \triangleBPQ,

                  PA = PB         (radii of same circle)

                 PQ = PQ        (Common)

                  QA = QB       (radii of same circle)

     So,           \triangleAPQ \cong \triangleBPQ       (By SSS)

                 \anglePAQ = \anglePBQ     (CPCT)

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seema garhwal

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