THEOREM * A theorem is a statement which has been proved to be true. A theorem is a proposition that has been or is to be proved on the basis of explicit assumptions. Proving theorems is a central activity of mathematicians. This meaning of the word "theorem" is distinct from the meanings of the word "theory". ...

REMAINDER THEOREM STATEMENT :

REMAINDER THEOREM STATEMENT Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number . If p(x) is divided by the linear polynomial x-a, then the remainder is p(a) .

REMAINDER THEOREM PROOF :

Suppose that when p(x) is divided by x-a the quotient is q(x) and the remainder is r(x) , i.e.,
p(x) = (x-a) q(x) + r(x)
Since x-a is a 1 & the degree of r(x) is less than the degree of x-a, degree of r(x) =0
Therefore p(x) = (x-a) q(x) + r
If x = a p(a) = (a-a) q(a) + r REMAINDER THEOREM PROOF

PYTHAGORAS THEOREM STATEMENT :

In a right angled triangle the square of the long side (the "hypotenuse") is equal to the sum of the squares of the other two sides.It is stated in this formula:a2 + b2 = c2 PYTHAGORAS THEOREM STATEMENT

PYTHAGORAS THEOREM PROOF :

PYTHAGORAS THEOREM PROOF a b c The square has a square hole with the side (a - b). Summing up its area (a - b)² and 2ab, the area of the four triangles (4·ab/2), we get

FACTOR THEOREM :

FACTOR THEOREM If p(x) is a polynomial of degree n > 1 & a is any real number then
x-a is a factor of p(x) , if p(a) = 0 &
p(a) = 0 , if x-a is a factor of p(x) By remainder theorem , p(x) = (x-a) p(x) + p(a)
If p(a) =0 then p(x) = 0 (x-a) p(x) , which shows that
x-a is a factor of p(x) . STATEMENT PROOF

CHAPTER - 6 LINES & ANGLES :

CHAPTER - 6 LINES & ANGLES Total Theorem - 8

THEOREM – 6.1 :

THEOREM – 6.1 If two lines intersect each othjer , then the vertically opposite angles are equal . In the statements above , it is given that ‘two lines intersect each other’ . So, let AB and CD be two lines intersecting at O . They lead to two pairs of vertically Opposite angles , namely, PROOF STATEMENT

THEOREM – 6.2STATEMENT :

THEOREM – 6.2STATEMENT If a transversal intersect two parallel lines , then each pair of alternate interior angles are equal .

THEOREM –6.2PROOF :

THEOREM –6.2PROOF The transversal PS intersects lines AB & CD at points Q & R respectively such that BQR = QRC BQR = PQA BQR = QRC So we may conclude that
PQA = QRC SO AB CD R P C S Q B D A

THEOREM – 6.3 :

THEOREM – 6.3 IF a transversal intersects two lines such that a pair of alternate interior angles is equal , then the two lines are parallel .

THEOREM – 6.4 :

THEOREM – 6.4 IF a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary .

THEOREM –6.5 :

THEOREM –6.5 IF a transversal intersects two lines, such that a pair of interior angles on the same side of the transversal is supplementary , then the two lines are parallel

THEOREM – 6.6 :

THEOREM – 6.6 Lines which are parallel to the same line are parallel to each other .

THEOREM – 6.7 STATEMENT :

THEOREM – 6.7 STATEMENT The sum of the angles of a triangle is 180 0 . P R Q R Q P Y X 2 1 3 4 3 2 1 5

THEOREM – 6.7 PROOF :

THEOREM – 6.7 PROOF The triangle PQR and angle 1,2,3 are angles of triangle PQR . A line XPY parallel to QR through the opposite vertex P . So that we can use the properties related to the parallel lines . Therefore , |_ 4 + |_ 1 + |_5 = 180 0 (1)
But XPY parallel to QR & PQ , PR are transversal.
|_2 + |_1 + |_3 = 180 0
so, |_1 + |_2 + |_3 = 180 0

THEOREM –6.8 :

THEOREM –6.8 If a side of a triangle is produced , then the exterior angle so formed is equal to the sum of the two interior opposite angles . An exterior angles of a triangle is greater than either of its interior opposite angles .

CHAPTER – 7 TRIANGLES :

CHAPTER – 7 TRIANGLES Total Theorem - 8

THEOREM – 7.1STATEMENT :

THEOREM – 7.1STATEMENT Two triangles are congruent if two angles & the included side of one triangle are equal to two angles and the included side of other triangle . ASA CONGRUENCE RULE

THEOREM – 7.1PROOF :

THEOREM – 7.1PROOF Let AB = DE
AB = DE
|_B = |_E
BC = EF
SO, ABC = DEF A B E C F D

THEOREM – 7.2STATEMENT :

THEOREM – 7.2STATEMENT Angle opposite to equal sides of an isosceles triangle are equal .

THEOREM – 7.2PROOF :

THEOREM – 7.2PROOF The bisector of |_A & let D be the point of intersection of this bisector of |_A & BC . IN BAD & CAD ,
AB = AC
|_BAD = |_CAD
AD = AD
So , BAD = CAD A B D C

THEOREM – 7.3STATEMENT :

THEOREM – 7.3STATEMENT The sides opposite to equal angles of a triangle are equal .

THEOREM – 7.3PROOF :

THEOREM – 7.3PROOF This theorem is Proved by ASA congruence rule . For example :- when In any
|_BAD = |_CAD
AD = AD
|_ADB = |_ADC = 90 0
So, ABD = ACD ( ASA Rule )
So, AB = AC ( CPCT ) A D C B

THEOREM – 7.4STATEMENT :

THEOREM – 7.4STATEMENT SSS CONGRUENCE RULE If three sides of one triangle are equal to the three sides of another triangle , then the two triangles are congruent.

THEOREM – 7.4PROOF :

THEOREM – 7.4PROOF SSS CONGRUENCE RULE This theorem can be proved by suitable construction A Q B R C P 4.5 cm 3.5 cm 4 cm 4.5 cm 4 cm 3.5 cm

THEOREM – 7.5 STATEMENT :

THEOREM – 7.5 STATEMENT RHS CONGRUENCE RULE If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle , then the two triangle are congruent .

THEOREM – 7.5 PROOF :

THEOREM – 7.5 PROOF RHS CONGRUENCE RULE The two triangles cover each other completely & so they are congruent . Two right triangles are congruent if one pair of sides & the hypotenuse are equal . 4 cm 4 cm 5 cm 5 cm

THEOREM – 7.6STATEMENT :

THEOREM – 7.6STATEMENT If two sides of a triangle are unequal , the angle opposite to the longer side is larger .

THEOREM – 7.6PROOF :

THEOREM – 7.6PROOF A Q R S T U B P This theorem can be proved by this diagram.

THEOREM – 7.7 :

THEOREM – 7.7 In any triangle , the side opposite to the larger angle is longer . STATEMENT Take a triangle ABC & find AB+BC, BC + AC, AC+AB Observation : AB+BC >AC , BC+AC >AB , And
AB+AC >BC . A B C PROOF

THEOREM – 7.8 :

THEOREM – 7.8 The sum of any two sides of a triangle is greater than the third side . PROOF D A B C This theorem is proved by this diagram STATEMENT

CHAPTER – 8 QUADRILATERAL :

CHAPTER – 8 QUADRILATERAL Total Theorem - 10

THEOREM – 8.1STATEMENT :

THEOREM – 8.1STATEMENT A diagonal of a parallelogram divides it into two congruent triangles .

THEOREM – 8.1PROOF :

THEOREM – 8.1PROOF Let ABCD be a parallelogram and AC be a diagonal . We will finddiagonal AC divides parallelogram ABCD into two triangles ABC & CDA . In triangle ABC & CDA BC//AD and AC is transversal
SO, |_BCA = |_DAC
AB = DC & AC is a transversal .
So, |_BCA = |_ DCA
and AC = CA
Therefore , Diagonal AC divides parallelogram ABCD into two congruent triangles ABC & CDA. D B A C

THEOREM – 8.2 :

THEOREM – 8.2 STATEMENT In a parallelogram , opposite sides are equal . PROOF In the previous theorem We have already learn that If a quadrilateral is a parallelogram then each pair of its opposite sides is equal . So opposite sides of parallelogram are equal . D B A C Therefore In this diagram AB = DC & AD = BC

THEOREM – 8.3 :

THEOREM – 8.3 If each pair of opposite sides of a quadrilateral is equal , then it is a parallelogram . STATEMENT PROOF Take quadrilateral ABCD & draw diagonal AC .We find that :
ABC = CDA
So, |_ DCA = |_ BAC
|_BCA = |_ DAC
and AB = CD , AD = BC D B A C

THEOREM – 8.4 :

THEOREM – 8.4 In a parallelogram , opposite angles are equal . STATEMENT PROOF This diagram is a proof of this theorem . D B A C

THEOREM – 8.5STATEMENT :

THEOREM – 8.5STATEMENT If in a quadrilateral , each pair of opposite angles is equal , then it is a parallelogram .

THEOREM – 8.6STATEMENT :

THEOREM – 8.6STATEMENT The diagonals of a parallelogram bisect each other . C A D B O

THEOREM – 8.7 :

THEOREM – 8.7 If the diagonal of a quadrilateral bisect each other then it is a parallelogram . PROOF OA=OC & OB=OD
So, AOB = COD
Therefore, |_ABO = |_CDO
So , AB || CD , BC || AD
Therefore ABCD is a parallelogram . C A D B O STATEMENT

THEOREM – 8.8 :

THEOREM – 8.8 A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel . PROOF In this fig. AB = CD and AB || CD
And ABC = CDA
So , BC || AD C A D B STATEMENT

THEOREM – 8.9 :

THEOREM – 8.9 The line segment joining the mid-points of two sides of a triangle is parallel to the third side . PROOF In this fig. E & F are mid points of AB & AC respectively and CD || BA
And AEF = CDF
So , EF || DF & BE = AE = DC
This gives EF || BC A F E D C B STATEMENT

THEOREM – 8.10 STATEMENT :

THEOREM – 8.10 STATEMENT The line drawn through the mid point of one side of a triangle parallel to another side bisects the third side . PROOF A F E D C B This diagram is a proof of this theorem .

CHAPTER – 9 AREAS OF PARALLELOGRAM & TRIANGLES :

CHAPTER – 9 AREAS OF PARALLELOGRAM & TRIANGLES Total Theorem - 3

THEOREM – 9.1 STATEMENT :

THEOREM – 9.1 STATEMENT Parallelogram on the same base and between the same parallels are equal in area .

THEOREM – 9.1 PROOF :

THEOREM – 9.1 PROOF Two parallelograms ABCD and EFCD , on the same base DC and between the same parallels AF and DC are given .
|_DAE = |_CBF ( corresponding |_ from AD || BC & trans. AF
|_AED = |_BFC ( corresponding |_ from AD || BC & trans.AF
Therefore, |_ADE = |_BCF ( Angle sum property of a triangle )
Also, AD = BC ( Oppst. Sides of the parallelogram ABCD )
So ADE = BCF ( ASA rule )
So, Parallelograms ABCD and EFCD
are equal in area . A F B D E A C

THEOREM – 9.2 STATEMENT :

THEOREM – 9.2 STATEMENT Two triangles on the same base ( or equal base ) and between the same parallels are equal in area .

THEOREM – 9.2 PROOF :

THEOREM – 9.2 PROOF A F B D E A C Suppose ABCD is a parallelogram whose one of the diagonal is AC . Let AN _|_ DC .
ADC = CBA
So ar(ADC) = ar (CBA)
Therefore, ar(ADC) = ½ ar (ABCD)
So, area of ADC = ½ base DC
corresponding altitude AN

THEOREM – 9.3 :

THEOREM – 9.3 Two triangles having the same base ( or equal bases ) and equal areas lie between the same parallels . STATEMENT

CHAPTER – 10 CIRCLES :

CHAPTER – 10 CIRCLES Total Theorem - 12

THEOREM – 10.1 STATEMENT :

THEOREM – 10.1 STATEMENT Equal chords of a circle subtend equal angles at the centre .

THEOREM – 10.1 PROOF :

THEOREM – 10.1 PROOF We have two equal chords AB and CD of a circle with centre O .
In triangle AOB and COD
OA = OC ( Radii of a circle )
OB = OD (Radii of a circle )
AB = CD
Therefore , AOB = COD ( SSS rule )
This gives |_AOB = |_COD ( Corresponding parts of congruent triangles ) O D B C A

THEOREM – 10.2 STATEMENT :

THEOREM – 10.2 STATEMENT If the angles subtended by the chords of a circle at the centre are equal , then the chords are equal . PROOF |_AOB = |_COD ( By above theorem )
AOB = COD
So, the chords AB & CD are equal . O B C A D

THEOREM – 10.3 STATEMENT :

THEOREM – 10.3 STATEMENT The perpendicular from the centre of a circle to a chord bisects the chord . O A M B

THEOREM – 10.4 STATEMENT :

THEOREM – 10.4 STATEMENT The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord .

THEOREM – 10.4 PROOF :

THEOREM – 10.4 PROOF O A M B Let AB be a chord of a circle with centre O and O is joined to the mid point M of AB . We have to prove that OM _|_ AB . Join OA and OB . In triangles OAM and OBM ,
OA = OB
AM = BM
OM = OM
Therefore OAM = OBM
This gives, |_OMA = |_OMB = 900

THEOREM – 10.5 :

THEOREM – 10.5 O Q There is one and only one circle passing through three given non-collinear points . R P

THEOREM – 10.6 :

THEOREM – 10.6 O M Equal chords of a circle ( or of congruent circles ) are equidistant from the centre ( or centres ) . L O M L S Q P R

THEOREM – 10.7 :

THEOREM – 10.7 Chords equidistant from the centre of a circle are equal in length . PROOF If two intersecting chords of a circle
make equal angles with the
diameter passing through
their point of intersection
so, the chords are equal . O D L M E Q A B P C STATEMENT

THEOREM – 10.8 STATEMENT :

THEOREM – 10.8 STATEMENT The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle .

THEOREM – 10.8 PROOF :

THEOREM – 10.8 PROOF Arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle . In all cases, |_BOQ = |_QAO + |_AQO
OA = OQ ( Radii of a circle )
Therefore, |_OAQ = |_OQA
So, |_BOQ = 2 |_OAQ , |_BOP = 2 |_OAP
By above |_BOP + |_BOQ = 2( |_OAP + |_OAQ )
|_POQ = 2 |_PAQ B P P Q A O O O B A B A Q Q P

THEOREM – 10.9 :

THEOREM – 10.9 Angles in the same segment of a circle are equal . A O Q C P We know that ,
|_PAQ = ½ |_POQ
|_PAQ = ½ 180 0 = 90 0
Same as |_PCQ which is equal to 900
So |_PAQ = |_PCQ PROOF

THEOREM – 10.10 STATEMENT :

THEOREM – 10.10 STATEMENT If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment , the four points lie on a circle ( i.e. they are concyclic )

THEOREM – 10.10 PROOF :

THEOREM – 10.10 PROOF C B E’ A AB is a line segment , which subtends equal angles at two point C & B . That is
|_ACB = |_ADB
If points A, C, E & B lie on circle .
|_ACB = |_AEB
|_ACB = |_ADB
Therefore |_AEB = |_ADB
So, this is not possible E coincides with D
Similarly E’ should also coincides with D E D

THEOREM – 10.11 :

THEOREM – 10.11 The sum of either pair of opposite angles of a cyclic quadrilateral is 1800 A C B D |_A + |_C = 1800
|_B + |_D = 1800

THEOREM – 10.12 :

THEOREM – 10.12 If the sum of a pair of opposite angles of a quadrilateral is 1800 , the quadrilateral is cyclic A C B D |_A + |_C = 1800
|_B + |_D = 1800
For that reason this quadrilateral is cylic

Slide 69:

PRESENTED BY :- HOPE YOU HAVE UNDERSTAND WELL NILESH DASHORE
CLASS - X

You do not have the permission to view this presentation. In order to view it, please
contact the author of the presentation.

Send to Blogs and Networks

Processing ....

Premium member

Use HTTPs

HTTPS (Hypertext Transfer Protocol Secure) is a protocol used by Web servers to transfer and display Web content securely. Most web browsers block content or generate a “mixed content” warning when users access web pages via HTTPS that contain embedded content loaded via HTTP. To prevent users from facing this, Use HTTPS option.

By: vibhutripathi184 (41 month(s) ago)

You need to sign-up