kpk-10th-maths-ch08-12-theorems.pdf - MathCity.org

Theorems

Khalid Mehmood Lecturer GDC Shah Essa Bilot Sharif Available at www.mathcity.org

124

CHPATER 8

CHPATER 8

Theorem 8.1 Statement: In an obtuse angled triangle, the square on the side opposite

to the obtuse angle is equal to the sum of the square on the sides containing the obtuse angle

together with twice the rectangle contained be the sides, and the projection on it of the other.

Given: In oABC,m C 90 , Let us denotemAB c,mAC b

and mAB c . mAD h is perpendicular from A to BC ( Produced

) so that mCD p is projection of AC on BC

To Prove: 2 2 2

mAB mAC mBC 2 mBC mCD

Or 2 2 2c a b 2ap

Proof Statements Reasons

ADC is right angled triangle

2 2 2

mAC mAD mCD 1

ADB is right angled triangle

2 2 2

2 2 2

2 2 2 2

2 2 2 2

2 2 2

mAB mAD mBD 2

mAB mAD mBC mCD

mAB mAD mBC mCD 2 mBC mCD

mAB mAD mCD mBC 2 mBC mCD


Pythagoras theorem

Pythagoras theorem

2 2 2x y x y 2xy

Using Equation (1)

Theorem 8.2: In any triangle, square on the side opposite to the acute angle is equal

to the sum of the squares on the sides containing that acute angle diminished by twice

the rectangle contained by one of those sides and the projection on it of the other.

Figure A is acute angled triangle Figure B is obtuse angle triangle

Given: In oABC,m C 90 , Let us denote mAB c,mAC b and mAB c . mAD h is

perpendicular from A to BC ( Produced ) so that mCD p is projection of AC on BC

To Prove: 2 2 2

mAB mBC mCA 2 mBC mCD Or 2 2 2c a b 2ap

Proof Statements Reasons ADC is right angled triangle

2 2 2

mAC mAD mCD 1


2 2 2

2 2 2

2 2 2 2

2 2 2 2

2 2 2

mAB mAD mBD 2

mAB mAD mBC mCD

mAB mAD mBC mCD 2 mBC mCD

mAB mAD mCD mBC 2 mBC mCD


Pythagoras theorem

Pythagoras theorem

mBC mBD mDC

mBC mDC mBD

2 2 2x y x y 2xy

Using Equation (1)

Theorems


125

CHPATER 8

Theorem 8.3 Statement: In any triangle, the sum of square of any two sides is equal

to twice square of on half the third side together with twice the square on median which bisect

the third side.

Figure 1 is acute angled triangle Figure 2 is obtuse angle triangle

Given: In ABC,AE , is the median drawn from the Vertex A then mBE mEC .

To Prove: 2 2 2 2

mAB mAC 2 mEB 2 mAE

Proof Statements Reasons ADE is right angled triangle

2 2 2

mAE mAD mED 1

ADC is right angled triangle

2 2 2

mAC mAD mCD 2


2 2 2

mAB mAD mBD 3

2 2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2

mAB mAC mAD mAD mCD mBD

mAB mAC 2 mAD mEC mED mBE mED

mAB mAC 2 mAD mBE mED mBE mED

mAB mAC 2 mAD 2 mBE 2 mED

mAB mAC 2 mAD 2 mED 2 mBE

mAB mAC 2 mAD mED 2 mBE

mAB mAC 2 m

2 2

AE 2 mBE

Pythagoras theorem

Pythagoras theorem

Pythagoras theorem Adding equations (2) and (3)

mDE mDC mEC

mDC mEC mDE

mBD mBE mED

mDC mEC mDE

Since mBE mEC

2 2 2 2x y x y 2x 2y

Using Equation (1)

Circle: A circle is a set of points in a plane which are equidistance from a fixed point of the plane.

The fixed point is called center of the circle.

Radius of the Circle: The radius of the circle is a

straight line drawn from the center to the boundary line or

circumference. The plural of the word radius is radii. OG

Circumference: The circumference of a circle is the

boundary line or perimeter of the circle. Chord: A line segment whose end points are any two points

of a circle is called a chord of the circle. AB

Diameter of the circle: A chord passing through

center of circle is called diameter of the circle. EF Tangent to a circle: A line which intersects a circle at one point only and none of its points lie in the

interior of circle is called a tangent to circle. XY

Secant to a circle: If there are two distinct points common between a line and a circle, the line

is called secant to the circle. If there no point common between a line and a circle, we say that

the line and the circle do not intersect.CD

Theorems


126

CHPATER 9

CHPATER 9

Theorem 9.1: One and only one circle can pass through three non-collinear points.

Given: A,B and C are three non-collinear points

To Prove: One and only one circle can pass through

the points A,B and C i.e., OA OB OC

Construction: Join B to A and C, Draw perpendicular

bisectors XD and YE of AB and BC respectively,

which intersect at O. Proof Statements Reasons

In OAD OBD

ODA ODB

AD BD

OD OD

OAD OBD

OA OB 1

Now in OBE OCE

OEB OEC

BE EC

OE OE

OBE OCE

OB OC 2

OA OB OC

mOA mOB mOC

XD AB (Construction)

XD is bisector of AB

Common

SAS Postulate Corresponding sides of congruent triangles

YE BC

YE is bisector of BC

Common


From equations (1) & (2)/Transitive

property

It means that O is the only one point which is equidistance from the three points A,B,C. Therefore, one

and only one circle with center O can pass through three non-collinear points i.e., radius

mOA mOB mOC

Theorem 9.2: A straight line, drawn from a centre of a circle to bisect a chord

(which is not diameter) is perpendicular to the chord

Given: A circle with centre O, AB is a chord of the circle,

N is the midpoint of AB which is joined to O

To Prove: ON AB

Construction: Join O to A and B. Proof Statements Reasons In OAN OBN

OA OB

AN BN

ON ON

OAN OBN

m 1 m 2

But om 1 m 2 180

o

o

o

o

m 1 m 1 180

2m 1 180

180m 1

2m 1 90

om 1 m 2 90

Hence ON AB

Radial segments

N is mid point of AB

Common

SSS Postulate Corresponding angles of congruent triangles

Supplementary angles postulate

Proved m 1 m 2

Since m 1 m 2

Theorems


127

CHPATER 9

Theorem 9.3: Perpendicular from the centre of a circle on a chord bisect it.

Given: A circle with centre O, AB is a chord and ON AB

To Prove: N is the mid point of AB

i.e., AN BN

Construction: Join O to A and B. Proof Statements Reasons In OAN OBN

OA OB

m ONA m ONB

ON ON

OAN OBN

AN BN

Radial segments

Given/Right angles

Common

HS Postulate Corresponding sides of congruent triangles

Th 9.4: If two chords of a circle are congruent then they will be equidistant from centre

Given: AB and CD are two congruent chords of the circle

with centre O. i.e., mAB mCD

To Prove: AB and CD are equidistant from the centre O.

Construction: Join O to A and C. Also draw

perpendiculars OE and OF on the given chords AB and

CD respectively Proof Statements Reasons mAE mEB and om OEA 90

mCF mFD and om OFC 90

mAB mCD

mAE mEB mCF mFD

mAE mAE mCF mCF

2mAE 2mCF

mAE mCF

In OAE OCF

OA OC

m OEA m OFC

AE CF

OAE OCF

mOE mOF

OE is perpendicular bisector of AB

OF is perpendicular bisector of CD Given that

mAE mEB mCF mFD

In the correspondence

Radial segments

Construction

Proved

HS Postulate

Corresponding sides of congruent triangles

Theo 9.5: If two chords of a circle which are equidistant from centre are congruent.

Given: AB and CD are chords of the circle are

equidistant from centre O.

To Prove: mAB mCD

Construction: Join O to A and C. Also draw

perpendicular bisectors OE and OF on the given

chords AB and CD respectively.

Proof Statements Reasons mAE mEB and om OEA 90

mCF mFD and om OFC 90 In OAE OCF

OA OC

mOE mOF

m OEA m OFC

OE is perpendicular bisector of AB

OF is perpendicular bisector of CD In the correspondence

Radial segments

Given

Construction

Theorems


128

CHPATER 10

OAE OCF

AE CF

mAE mCF

2mAE 2mCF

mAE mAE mCF mCF

mAE mEB mCF mFD

mAB mCD

HS Postulate


Multiply both sides by 2

mAE mEB mCF mFD

CHPATER 10 Theorem 10.1: If a line is drawn perpendicular to a radial segment of a circle at its

outer end point, it is tangent to the circle at that point

Given: OC is radial segment of the circle with centre O.

Perpendicular line AB at outer end point of the circle C.

i.e., AB OC

To Prove: AB is tangent to the circle.

Construction: Take any point D on AB other than C Join O to D. Proof Statements Reasons

OC AB So the OCD is right angled triangle

m 1 and m 2 are acute angles

Thus m C m 2

mOD mOC

Given om C 90

i.e., om 1 m 2 90

m C is right and m 2 is acute In a triangle, greater angle has greater opposite side

mOD is greater then radial segment mOC Thus D is exterior of circle

Hence meets the circle at only one point which is C.

Theorem 10.2: The tangent to a circle and the radial segment joining the point of

contact and the centre are perpendicular to each other.

Given: AB is tangent to the circle at point C, OC is radial segment

which is obtained by joining the Centre O and the point of contact of the

tangent AB

To Prove: AB OC

Construction: Take any point D on AB other than C Join O to D. Proof Statements Reasons C is the only point common to the circle

and the tangent AB D is an exterior point of the circle

It means, mOD mOC

mOC is shortest distance between

the centre O and the line AB

AB OC

AB is tangent to the circle at point C

Construction (Except C every point of AB is

outside the circle)

m C is right (In a triangle greater angle has

greater side opposite to it)

By definition of shortest distance

T 10.3: Two tangents drawn to a circle from a point outside it, are equal in length.

Given: A circle with centre O. A is any point

outside the circle AB and AC are two tangents

from a point A.

To Prove: mAB mAC

Construction: Join O to A,B and C.

Theorems


129

CHPATER 10


AOB AOC

AO AO

OB OC

ABO ACO

AOB AOC

AB AC

Common

Radial segments

Right angles

HS HS Corresponding sides of congruent triangles

Theorem 10.4 Statement: The two circles touches externally, the distance

between their centres is equal to sum of their radii.

Given: Two circles with centre O and O’ which is touching

each other externally at point A.OA and O'A are the radial

segments of the circles

To Prove: mOO' mOA mAO'

Construction: Draw a common tangent BC at the

point A which is common point of the two circles Proof Statements Reasons

om BAO 90 1

om BAO' 90 2 o o

o

m BAO m BAO' 90 90

m BAO m BAO' 180

OA and AO' are opposite rays

Thus O, A and O’ are three

different collinear points

mOO' mOA mAO'

BC is tangent to a circle at point A i.e., OA BC

BC is tangent to a circle at point A i.e., O'A BC

m BAO and m BAO' are supplementary angles

with common vertex A

Straight line Postulate

Theorem 10.4: The two circles touches

internally, the distance between their centre is the

difference between their radii.

Given: Two circles with centre O and O’ which is touching

each other internally at point A.OA and O'A are the radial

segments of the circles

To Prove: mOO' mOA mO'A

Construction: Draw a common tangent BC at the

point A which is common point of the two circles Proof Statements Reasons

om BAO 90 1

om BAO' 90 2 om BAO m BAO' 90

Thus O, A and O’ are three

different collinear points

mOA mOO' mO'A

mOA mO'A mOO'

mOO' mOA mO'A

BC is tangent to a circle at point A i.e., OA BC

BC is tangent to a circle at point A i.e., O'A BC

m BAO and m BAO' are right angles with

common vertex A

A,O and O’ lies on the same straight line

Straight line Postulate

From the law of equation

Arc of the Circle: Any portion or part of the circle is called arc of the circle.

Major Arc: An arc which is greater than a semi-circle is called Major arc. Central Angle: An angle subtended by an arc at the centre of a circle is called central angle of arc.

Theorems


130

CHAPTER 11

CHAPTER 11 Theorem 11.1 (a): If two arcs of a circle are

congruent then the corresponding chords are equal

Given: A circle with centre O AEB and CFD are

congruent arcs. i.e., AEB CFD

To Prove: mAB mCD

Construction: Join O to A,B,C and D respectively

and label the central angles 1 and 2 Proof Statements Reasons In OAB OCD

OA OC

m 1 m 2

OB OD

OAB OCD

AB CD


Radial segments

Central angle of two congruent arcs

Radial segments


Theorem 11.1 (b): If two arcs of the congruent circles are congruent then the

corresponding chords are equal

Given: Two congruent circles with centre O and O’ respectively. AEB and CFD are congruent

arcs. i.e., AEB CFD

To Prove: mAB mCD

Construction: Join O to A,B and O’ to

C and D respectively and label the

central angles 1 and 2

Proof Statements Reasons In OAB O'CD

OA O'C

m 1 m 2

OB O'D

OAB O'CD

AB CD

Radii of congruent circles



SAS Postulate


Theorem 11.2 (a) If two chords of a circle are

congruent then the corresponding minor arcs are congruent

Given: A circle with centre O AB and CD are congruent chords. i.e.,

mAB mCD

To Prove: AEB CFD

Construction: Join O to A,B,C and D respectively and label

the central angles 1 and 2

Proof Statements Reasons In OAB OCD

OA OC

AB CD

OB OD

OAB OCD

m 1 m 2

AEB CFD


Radial segments

Given

Radial segments

SSS Postulate

Corresponding angles of congruent triangles

Congruent arcs has congruent Central angles

Theorems


131

CHAPTER 11

Theorem 11.2 (b): If two chords of a circle are congruent

then corresponding major arcs are congruent

Given: A circle with centre O AB and CD are congruent chords. i.e.,

mAB mCD

To Prove: AFB CED

Construction: Join O to A,B,C and D respectively and label

the central angles 1 and 2


OA OC

AB CD

OB OD

OAB OCD

m 1 m 2

AEB CFD

m AFB 360o - m AEB …………(1)

m CED 360o - m CFD …………(2)

mAFB mCED


Radial segments

Given

Radial segments

SSS Postulate

Corresponding angles of congruent triangles

Congruent arcs has congruent Central angles Total angle at a point = 360o

i.e., m AEB + m AFB=360o & m CFD + m CED=360o

Transitive property

Theorem 11.2 (c) Statement: If two chords of the congruent circles are

congruent then the corresponding minor arcs are congruent

Given: Two congruent circles with centre O

and O’ respectively. AB and CD are congruent

chords. i.e., mAB mCD

To Prove: AEB CFD

Construction: Join O to A,B and O’ to C

and D respectively and label the central

angles 1 and 2


OA O'C

AB CD

OB O'D

OAB O'CD

m 1 m 2

mAEB mCFD



Given


SSS Postulate

Corresponding angle of congruent triangles


Theorem 11.2 (d) Statement: If two chords of the congruent circles are

congruent then the corresponding major arcs are congruent

Given: Two congruent circles with

centre O and O’ respectively. AB and

CD are congruent chords. i.e.,

mAB mCD

To Prove: AGB CHD

Construction: Join O to A,B and O’

to C and D respectively and label the



Theorems


132

CHAPTER 11

In OAB O'CD

OA O'C

AB CD

OB O'D

OAB O'CD

m 1 m 2

mAEB mCFD

omAGB 360 mAEB …………(1)

omCHD 360 mCFD …………(2)

mAGB mCHD



Given


SSS Postulate Corresponding angle of congruent triangles


Total angle at a point = 360o

i.e., m AEB + m AFB=360o & m CFD + m CED=360o

Transitive property

Theorem 11.2 (e): If two chords (diameters)

of a circle are congruent then the corresponding

minor/major arcs are congruent

Given: A circle with centre O AB and CD are diameters.

i.e., mAB mCD

To Prove: AFB AEB CED CFD Proof Statements Reasons

mAB mCD Then length of minor arc = length major arc

i.e., OmAEB mAFB 180 ………(1)

similarly OmCFD CED 180 …..……(2)

mAFB mAEB mCED mCFD

Diameters of the same circles

Congruent arcs has congruent Central angles Diameter is a straight line which has supplementary

angle

Transitive property

Theorem 11.2 (e): If two chords (diameters) of two congruent circles are congruent then corresponding minor/major arcs are congruent

Given: Two circles with centre O and O’. AB

and CD are diameters. i.e., mAB mCD

To Prove: AGB AEB CHD CFD Proof Statements Reasons

mAB mCD Then length of minor arc = length major arc

i.e., oAGB AEB 180 ………(1)

similarly oCFD CHD 180 ………(2)

mAGB mAEB mCHD mCFD

Diameters of the congruent circles

Congruent arcs has congruent Central angles Diameter is a straight line which has supplementary angle

Transitive property

Theorem 11.3 (a): Equal chords of a circle subtend

equal angles at the centre

Given: A circle with centre O and mAB mCD

To Prove: m 1 m 2

Construction: Join O to A,B,C and D respectively and

label the central angles 1 and 2 Proof Statements Reasons In OAB OCD

OA OC

AB CD

OB OD

OAB OCD

m 1 m 2

Radial segments

Given

Radial segments

SSS Postulate Corresponding angles of congruent triangles

Theorems


133

CHAPTER 11

Theorem 11.3 (b): Equal chords of congruent circles subtend equal angles at their centres


and O’ respectively. AB and CD are

congruent chords. i.e., mAB mCD

To Prove: m 1 m 2


C and D respectively and label the



OA O'C

AB CD

OB O'D

OAB O'CD

m 1 m 2


Given


SSS Postulate

Corresponding angle of congruent triangles

Theorem 11.4 (a) Statement: If the angles

subtend by two chords of circle at the centre are equal in

measures, then the chords are equal in measure

Given: A circle with centre O AB and CD are chords of

circle and m 1 m 2

To Prove: mAB mCD

Construction: Join O to A,B,C and D respectively and

label the central angles 1 and 2


OA OC

m 1 m 2

OB OD

OAB OCD

AB CD

Radial segments

Given

Radial segments

SAS Postulate


Theorem 11.4 (b): If the angles subtend by two chords of congruent circle at

the centre are equal in measures, then the chords are equal in measure


and O’ respectively. AB and CD are chords

of circle and m 1 m 2

To Prove: mAB mCD


C and D respectively


OA O'C

m 1 m 2

OB O'D

OAB O'CD

AB CD





SAS Postulate


Theorems


134

CHAPTER 12

CHAPTER 12

Theorem 12.1 (a): The measure of a central angle of a minor arc of a circle, is

double that of the angle subtended by the corresponding major arc

Given: A circle with centre O, AQB is minor arc whose

central angle is AOB . P is any point on the major arc ADB.

APB is the angle subtended by the arc at P.

To Prove: m AOB 2m APB

Construction: Draw PO to meet circumference at Q Proof Statements Reasons In OAP

OA OP

m OAP m OPA

m AOQ m OAP m OPA

m AOQ m OAP m OAP

m AOQ 2m OAP 1

In OPB

OB OP

m OAB m OPB

m BOQ m OBP m OPB

m BOQ m OPB m OPB

m BOQ m OPB 2

m AOQ m BOQ 2m OPA 2m OPB

m AOB 2 m OPA m OPB

m AOB 2m APB

Radial segments of circle

Congruent sides has congruent angles(Isosceles)

Exterior angle = sum of opposite interior angles

m OAP m OPA




m OBP m OPB

Adding equation (1) and (2)

Theorem 12.1 (b): Measure of a central angle of a minor arc of a circle(semi

circle), is double that of the angle subtended by the corresponding major arc

Given: A circle with centre O, AQB is minor arc whose

central angle is AOB . P is any point on the major arc ADB.

APB is the angle subtended by the arc at P.

To Prove: m AOB 2m APB

Construction: Draw PO to meet circumference at Q

Proof Statements Reasons In OAP

OA OP

m OAP m OPA

m AOQ m OAP m OPA

m AOQ m OAP m OAP

m AOQ 2m OAP 1

In OPB

OB OP

m OAB m OPB

m BOQ m OBP m OPB

m BOQ m OPB m OPB

m BOQ m OPB 2

m BOQ m AOQ 2m OPB 2m OPA

m AOB 2 m OPA m OPB

m AOB 2m APB




m OAP m OPA




m OBP m OPB

Subtracting equation (1) from (2)

Theorems


135

CHAPTER 12

Theorem 12.2 : Two angles in the same segments of a circle are equal

Given: A circle with centre O, ACB and ADB are any two angles in same arc ACDB containing

segments AB

To Prove: ACB ADB

Construction: Join O to A and B


m AOB 2m ACB 1

m AOB 2m ADB 2

2m ACB 2m ADB m ACB m ADB

Central angle = 2 ( angle subtended by major arc)


From equation (1) and (2)

Halves of equal quantities is again equal

Theorem 12.3 (a) Statement: The angle in a

semi-circle is a right angle.

Given: A circle with centre O, AB is a diameter and ACB is

any angle in the semi-circle.

To Prove: om ACB 90

Construction: Join O to C

Proof Statements Reasons In OAC

OA OC

m OAC m OCA 1

In OCB

OB OC

m OBC m OCB 2

m OAC m OBC m OCA m OCB

m OAC m OBC m ACB

o

o

o

o

But m OAC m OBC m ACB 180

m ACB m ACB 180

2m ACB 180

m ACB 90





Adding equation (1) and (2)

Sum of interior angles equals to 180o

m OAC m OBC m ACB

Again (Second Method) Theorem 12.3 (a) Statement: The angle in a semi-circle is a right angle.

Given: A circle with centre O, AB is a diameter and ACB is

any angle in the semi-circle.

To Prove: om ACB 90


o

o

o

m AOB 2m ACB

180 2m ACB

180m ACB

290 m ACB


∴ om AOB 180

Theorems


136

Theorem 12.3 (b) Angle in a segment greater then a semi-circle is less than a right angle

Given: A circle with centre O, AB is a chord below the

diameter EF is a diameter Thus the arc AECDFB if greater

than the semi-circular region.

To Prove: om ADB 90

Construction: Join D to E and F such that EDF is in

the semi-circle


om EDF 90

o

o

m EDA m ADB m BDF 90

m ADB 90 m EDA m BDF

om ADB 90

Measure of angle in semicircle is right

m EDF m EDA m ADB m BDF

Whole is greater than its parts

Theorem 12.3 (C)

Statement: The angle in a segment less than a semi-circle is greater than a right angle

Given: A circle with centre O, AB is a chord above the

diameter EF is a diameter Thus the arc ACDB if less than the

semi-circular region.

To Prove: om ADB 90

Construction: Join D to E and F such that EDF is in

the semi-circle Proof Statements Reasons

om EDF 90

o

o

o

m ADE m EDF m FDB m ADB

m ADE 90 m FDB m ADB

90 m ADB

m ADB 90

Measure of angle in semi-circle is right

m ADB m ADE m EDF m FDB

Whole is greater than its parts

Theorem 12.4: The opposite angles of any

quadrilateral inscribed in a circle are supplementary.

Given: A circle with centre O, ABCD is quadrilateral

inscribed in a circle

To Prove: o

o

m BAD m BCD 180

m ADC m ABC 180

Construction: Join O to B and D and label the angles

such that their sum om 1 m 2 360


2m BCD m 1 1

2m DAB m 2 2

o

o

o

2m BCD 2m DAB m 1 m 2

2 m BCD m DAB 360

360m BCD m DAB

2m BCD m DAB 180



Adding equations (1) and (2) om 1 m 2 360

Halves of equal quantities is again equal

Similarly we can prove that om ADC m ABC 180

kpk-10th-maths-ch08-12-theorems.pdf - MathCity.org

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