Q&A - Ask Doubts and Get Answers

#10.6

Q : 10 In any triangle ABC, if the angle bisector of $\small \angle A$ 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 $\small \angle A$ and perpendicular bisector of BC intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let $\angle$ ABD = $\angle$ 1 , $\angle$ ADC = $\angle$ 2 , $\angle$ DCB = $\angle$ 3 , $\angle$ CBD = $\angle$ 4

$\angle$ 1 and $\angle$ 3 lies in same segment.So,

$\angle$ 1 = $\angle$ 3 ..........................1(angles in same segment)

similarly, $\angle$ 2 = $\angle$ 4 ......................2

also, $\angle$ 1= $\angle$ 2 ..............3(given)

From 1,2,3 , we get

$\angle$ 3 = $\angle$ 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.

View Full Answer(1)

Posted by

seema garhwal

#10.6

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 $\small BP=BQ$ .

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.

$\therefore \angle APB = \angle AQB$

In $\triangle BPQ,$

$\angle APB = \angle AQB$

$\therefore BQ = BP$ (Sides opposite to equal of the triangle are equal )

View Full Answer(1)

Posted by

mansi

#10.6

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

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

Proof :

$\angle$ 1 and $\angle$ 3 are angles in same segment.therefore,

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

and $\angle$ 2 = $\angle$ 4 ..................2

Adding 1 and 2,we have

$\angle$ 1+ $\angle$ 2= $\angle$ 3+ $\angle$ 4

$\Rightarrow \angle D=\frac{1}{2}\angle B+\frac{1}{2}\angle C$ ,

$\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)$

$\Rightarrow \angle D=\frac{1}{2}(180 \degree+\angle C)$

and $\Rightarrow \angle D=\frac{1}{2}(180 \degree-\angle A)$

$\Rightarrow \angle D=90 \degree-\frac{1}{2}\angle A$

Similarly, $\Rightarrow \angle E=90 \degree-\frac{1}{2}\angle B$ and $\angle F=90 \degree-\frac{1}{2}\angle C$

View Full Answer(1)

Posted by

seema garhwal

#10.6

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

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

To prove : AE = AD

Proof :

$\angle$ ADC = $\angle$ 3 , $\angle$ ABC = $\angle$ 4, $\angle$ ADE = $\angle$ 1 and $\angle$ AED = $\angle$ 2

$\angle 3+\angle 1=180 \degree$ .................1(linear pair)

$\angle 2+\angle 4=180 \degree$ ....................2(sum of opposite angles of cyclic quadrilateral)

$\angle$ 3 = $\angle$ 4 (oppsoite angles of parallelogram )

From 1 and 2,

$\angle$ 3+ $\angle$ 1 = $\angle$ 2 + $\angle$ 4

From 3, $\angle$ 1 = $\angle$ 2

From 4, $\triangle$ AQB, $\angle$ 1 = $\angle$ 2

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

View Full Answer(1)

Posted by

mansi

#10.6

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 $\small \angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Given : AD = CE

To prove : $\angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)$

Construction: Join AC and DE.

Proof :

Let $\angle$ ADC = x , $\angle$ DOE = y and $\angle$ AOD = z

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

$\angle$ AOC + $\angle$ DOE + $\angle$ AOD + $\angle$ EOC = $360 \degree$

$\Rightarrow x+y+z+z=360 \degree$

$\Rightarrow x+y+2z=360 \degree$ .........................................1

In $\triangle$ OAD ,

OA = OD (Radii of the circle)

$\angle$ OAD = $\angle$ ODA (angles opposite to equal sides )

$\angle$ OAD + $\angle$ ODA + $\angle$ AOD = $180 \degree$

$\Rightarrow 2\angle OAD+z=180 \degree$

$\Rightarrow 2\angle OAD=180 \degree-z$

$\Rightarrow \angle OAD=\frac{180 \degree-z}{2}$

$\Rightarrow \angle OAD=90 \degree-\frac{z}{2}$ .............................................................2

Similarly,

$\Rightarrow \angle OCE=90 \degree-\frac{x}{2}$ .............................................................3

$\Rightarrow \angle OED=90 \degree-\frac{y}{2}$ ..............................................................4

$\angle$ ODB is exterior of triangle OAD . So,

$\angle$ ODB = $\angle$ OAD + $\angle$ ODA

$\Rightarrow \angle ODB=90 \degree-\frac{z}{2}+z$ (from 2)

$\Rightarrow \angle ODB=90 \degree+\frac{z}{2}$ .................................................................5

similarly,

$\angle$ OBE is exterior of triangle OCE . So,

$\angle$ OBE = $\angle$ OCE + $\angle$ OEC

$\Rightarrow \angle OEB=90 \degree-\frac{z}{2}+z$ (from 3)

$\Rightarrow \angle OEB=90 \degree+\frac{z}{2}$ .................................................................6

From 4,5,6 ;we get

$\angle$ BDE = $\angle$ BED = $\angle$ OEB - $\angle$ OED

$\Rightarrow \angle BDE=\angle BED=90 \degree+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}$

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

In $\triangle$ BDE ,

$\angle$ DBE + $\angle$ BDE + $\angle$ BED = $180 \degree$

$\Rightarrow \angle DBE +y+z=180 \degree$

$\Rightarrow \angle DBE =180 \degree-(y+z)$

$\Rightarrow \angle ABC =180 \degree-(y+z)$ ...................................................8

Here, from equation 1,

$\frac{x-y}{2}=\frac{360 \degree-y-2x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360 \degree-2y-2x}{2}$

$\Rightarrow \frac{x-y}{2}=180 \degree-y-x$ ...................................9

From 8 and 9,we have

$\angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)$

View Full Answer(1)

Posted by

mansi

#10.6

Q : 3 The lengths of two parallel chords of a circle are $\small 6\hspace {1mm}cm$ and $\small 8\hspace {1mm}cm$ . If the smaller chord is at distance $\small 4\hspace {1mm}cm$ 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 $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle and $OM \perp CD$

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 $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$ .............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$ .............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$ (OC=OA =radii)

$\Rightarrow 3 ^2+4^2=4^2+(4-x)^2$

$\Rightarrow 9+16=16+16+x^2-8x$

$\Rightarrow 9=16+x^2-8x$

$\Rightarrow x^2-8x+7=0$

$\Rightarrow x^2-7x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow (x-1)(x-7)=0$

$\Rightarrow x=1,7$

So, x=1 (since $x\neq 7> OM$ )

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

Hence, second chord is 3 cm away from centre.

View Full Answer(1)

Posted by

seema garhwal

#10.6

Q: 2 Two chords AB and CD of lengths $\small 5\hspace {1mm}cm$ and $\small 11\hspace {1mm}cm$ 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 $\small 6\hspace {1mm}cm$ , find the radius of the circle.

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

To find Radius (OA).

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle and $OM \perp CD$

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 $\triangle$ OCM , using pythagoras,

$OC ^2=CM^2+OM^2$ .............................1

and

In $\triangle$ OAN , using pythagoras,

$OA ^2=AN^2+ON^2$ .............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$ (OC=OA =radii)

$5.5 ^2+x^2=2.5^2+(6-x)^2$

$\Rightarrow 30.25+x^2=6.25+36+x^2-12x$

$\Rightarrow 30.25-42.25=-12x$

$\Rightarrow -12=-12x$

$\Rightarrow x=1$

From 2, we get

$OC^2=5.5^2+1^2=30.25+1=31.25$

$\Rightarrow OC=\frac{5}{2}\sqrt{5} cm$

OA = OC

Thus, the radius of the circle is $\frac{5}{2}\sqrt{5} cm$

View Full Answer(1)

Posted by

mansi

#10.6

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 $90 \degree$ , is a rectangle )

Thus, ABCD is rectangle.

View Full Answer(1)

Posted by

mansi

#10.6

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 $\triangle$ ABD and $\triangle$ CDO,

AO = OC (Given )

$\angle$ AOB = $\angle$ COD (Vertically opposite angles )

BO = DO (Given )

So, $\triangle$ ABD $\cong$ $\triangle$ CDO (By SAS)

$\angle$ BAO = $\angle$ DCO (CPCT)

$\angle$ BAO and $\angle$ DCO are alternate angle and are equal .

So, AB || DC ..............1

Also AD || BC ...............2

From 1 and 2,

$\angle A+\angle C=180 \degree$ ......................3(sum of opposite angles)

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

From 3 and 4,

$\angle A+\angle A=180 \degree$

$\Rightarrow 2\angle A=180 \degree$

$\Rightarrow \angle A=90 \degree$

BD is a diameter of the circle.

Similarly, AC is a diameter.

View Full Answer(1)

Posted by

seema garhwal

#10.6

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 : $\angle$ PAQ = $\angle$ PBQ

Proof : In $\triangle$ APQ and $\triangle$ BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, $\triangle$ APQ $\cong$ $\triangle$ BPQ (By SSS)

$\angle$ PAQ = $\angle$ PBQ (CPCT)

View Full Answer(1)

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

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.

Posted by

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 .

Posted by

mansi

Crack CUET with india's "Best Teachers"

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

Posted by

seema garhwal

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

Posted by

mansi

JEE Main high-scoring chapters and topics

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.

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

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 $\small BP=BQ$ .

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

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

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 $\small \angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Q : 3 The lengths of two parallel chords of a circle are $\small 6\hspace {1mm}cm$ and $\small 8\hspace {1mm}cm$ . If the smaller chord is at distance $\small 4\hspace {1mm}cm$ from the centre, what is the distance of the other chord from the centre?

Q: 2 Two chords AB and CD of lengths $\small 5\hspace {1mm}cm$ and $\small 11\hspace {1mm}cm$ 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 $\small 6\hspace {1mm}cm$ , find the radius of the circle.

Q : 7 AC and BD are chords of a circle which bisect each other. Prove that

(ii) ABCD is a rectangle.

Q: 7 AC and BD are chords of a circle which bisect each other. Prove that

(i) AC and BD are diameters