Slide 1: INTRODUCTION NAME = DEEPAK SHARMA
CLASS = IX - B
SUBJECT = MATHS
SCHOOL = KAPIL GYANPEETH
Slide 2: TRIANGLES A Triangle is formed by three lines segments obtained by joining three pairs of points taken from a set of three non collinear points in the plane.
In fig 1.1 , three non collinear points A, B, C have been joined and the figure ABC, enclosed by three line segments , AB,BC, and CA is called a triangle.
The symbol ▲ (delta) is used to denote a triangle. The three given points are called the vertices of a triangle. The three line segments are called the sides of a triangle. The angle made by the line segment at the vertices are called the angles of a triangle.
If the sides of a ▲ABC are extended in order, then the angle between the extended and the adjoining side is called the exterior angles of the triangle.
In the fig. 1.2
1 ,2 and 3 are the exterior angles while 4 ,5,and 6 are the interior angles of the triangles. 1 4 5 2 6 3 A B B C A C B Fig. 1.1 Fig. 1.2
Slide 3: CLASSIFICATION ON THE BASIS OF SIDES
SCALENE TRIANGLE – If all the sides of triangle are unequal then it is a scalene tirangle.
ISOSCELES TRIANGLE –If any two sides of a triangle are equal then it is a isosceles triangle.
EQUILATERAL TRIANGLE- If all the sides of a triangle are equal then it is an equilateral triangle.
ON THE BASIS OF ANGLES
ACUTE ANGLED TRIANGLE –If all the three angles of a triangle are less than 900 then it is an acute anglesd triangle.
RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 900 then it is a right angled triangle.
OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 900 then it is a obtuse angled triangle.
Slide 4: CONGRUENCE OF TRIANGLES Two triangles are congruent if three sides and three angles of one triangle are equal to the corresponding sides and angles of other triangle.
The congruence of two triangles follows immediately from the congruence of three lines segments and three angles.
Slide 5: ANXIOMS SIDE – ANGLE – SIDE RULE (SAS RULE)
Two triangles are congruent if any two sides and the includes angle of one triangle is equal to the two sides and the included angle of other triangle.
EXAMPLE :- (in fig 1.3)
GIVEN: AB=DE, BC=EF ,
B= E
SOLUTION: IF AB=DE, BC=EF , B= E then by SAS Rule
▲ABS = ▲DEF 4 cm 4 cm 600 600 A B C D F E Fig. 1.3
Slide 6: ANGLE – SIDE – ANGLE RULE (ASA RULE )
Two triangles are congruent if any two angles and the included side of one triangle is equal to the two angles and the included side of the other triangle.
EXAMPLE : (in fig. 1.4)
GIVEN: ABC= DEF,
ACB= DFE,
BC = EF
TO PROVE : ▲ABC = ▲DEF
ABC = DEF, (GIVEN)
ACB = DFE, (GIVEN)
BS = EF (GIVEN)
▲ABC = ▲DEF (BY ASA RULE) A B C D E F Fig. 1.4
Slide 7: 3. ANGLE – ANGLE – SIDE RULE (AAS RULE)
Two triangles are congruent if two angles and a side of one triangle is equal to the two angles and one a side of the other.
EXAMPLE: (in fig. 1.5)
GIVEN: IN ▲ ABC & ▲DEF
B = E
A= D
BC = EF
TO PROVE :▲ABC = ▲DEF
B = E
A = D
BC = EF
▲ABC = ▲DEF (BY AAS RULE) D E F A B C Fig. 1.5
Slide 8: SSS RULE 4. SIDE – SIDE – SIDE RULE (SSS RULE)
Two triangles are congruent if all the three sides of one triangle are equal to the three sides of other triangle.
Example: (in fig. 1.6)
Given: IN ▲ ABC & ▲DEF
AB = DE , BC = EF , AC = DF
TO PROVE : ▲ABC = ▲DEF
AB = DE (GIVEN )
BC = EF (GIVEN )
AC = DF (GIVEN )
▲ABC = ▲DEF (BY SSS RULE) D E F A B C Fig. 1.6
Slide 9: 5. RIGHT – HYPOTENUSE – SIDE RULE (RHS RULE )
Two triangles are congruent if the hypotenuse and the side of one triangle are equal to the hypotenuse and the side of other triangle.
EXAMPLE : (in fig 1.7)
GIVEN: IN ▲ ABC & ▲DEF
B = E = 900 , AC = DF , AB = DE
TO PROVE : ▲ABC = ▲DEF
B = E = 900 (GIVEN)
AC = DF (GIVEN)
AB = DE (GIVEN)
▲ABC = ▲DEF (BY RHS RULE) D E F A B C 900 900 Fig. 1.7
Slide 10: SOME PROPERTIES OF TRIANGLE The angles opposite to equal sides are always equal.
Example: (in fig 1.8)
Given: ▲ABC is an isosceles triangle in which AB = AC
TO PROVE: B = C
CONSTRUCTION : Draw AD bisector of BAC which meets BC at D
PROOF: IN ▲ABC & ▲ACD
AB = AD (GIVEN)
BAD = CAD (GIVEN)
AD = AD (COMMON)
▲ABD = ▲ ACD (BY SAS RULE)
B = C (BY CPCT) A B D C Fig. 1.8
Slide 11: 2. The sides opposite to equal angles of a triangle are always equal.
Example : (in fig. 1.9)
Given : ▲ ABC is an isosceles triangle in which B = C
TO PROVE: AB = AC
CONSTRUCTION : Draw AD the bisector of BAC which meets BC at D
Proof : IN ▲ ABD & ▲ ACD
B = C (GIVEN)
AD = AD (GIVEN)
BAD = CAD (GIVEN)
▲ ABD = ▲ ACD (BY ASA RULE)
AB = AC (BY CPCT) A B D C Fig. 1.9
Slide 12: INEQUALITIES When two quantities are unequal then on comparing these quantities we obtain a relation between their measures called “ inequality “ relation.
Slide 13: Theorem 1 . If two sides of a triangle are unequal the larger side has the greater angle opposite to it. Example: (in fig. 2.1)
Given : IN ▲ABC , AB>AC
TO PROVE : C = B
Draw a line segment CD from vertex such that AC = AD
Proof : IN ▲ACD , AC = AD
ACD = ADC --- (1)
But ADC is an exterior angle of ▲BDC
ADC > B --- (2)
From (1) &(2)
ACD > B --- (3)
ACB > ACD ---4
From (3) & (4)
ACB > ACD > B , ACB > B ,
C > B A B D C Fig. 2.1
Slide 14: THEOREM 2. In a triangle the greater angle has a large side opposite to it
Example: (in fig. 2.2)
Given: IN ▲ ABC B > C
TO PROVE : AC > AB
PROOF : We have the three possibility for sides AB and AC of ▲ABC
AC = AB
If AC = AB then opposite angles of the equal sides are equal than
B = C
AC ≠ AB
(ii) If AC < AB
We know that larger side has greater angles opposite to it.
AC < AB , C > B
AC is not greater then AB
If AC > AB
We have left only this possibility AC > AB A C B Fig. 2.2
Slide 15: THEOREM 3. The sum of any two angles is greater than its third side
Example (in fig. 2.3) TO PROVE : AB + BC > AC
BC + AC > AB
AC + AB > BC
CONSTRUCTION: Produce BA to D such that AD + AC .
Proof: AD = AC (GIVEN)
ACD = ADC (Angles opposite to equal sides are equal )
ACD = ADC --- (1)
BCD > ACD ----(2)
From (1) & (2) BCD > ADC = BDC
BD > AC (Greater angles have larger opposite sides )
BA + AD > BC ( BD = BA + AD)
BA + AC > BC (By construction)
AB + BC > AC
BC + AC >AB A C B D Fig. 2.3
Slide 16: THEOREM 4. Of all the line segments that can be drawn to a given line from an external point , the perpendicular line segment is the shortest.
Example: (in fig 2.4)
Given : A line AB and an external point. Join CD and draw CE AB
TO PROVE CE < CD
PROOF : IN ▲CED, CED = 900
THEN CDE < CED
CD < CE ( Greater angles have larger side opposite to them. ) B A C E D Fig. 2.4
Slide 17: SOME OTHER APPLICATIONS OF ASA AND AAS CONGRUENCE CRITERIA If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles triangle.Example : (in fig.2.5)
Given : A ▲ABC such that the altitude AD from A on the opposite side BC bisects BC i. e. BD = DC
To prove : AB = AC
SOLUTION : IN ▲ ADB & ▲ADC
BD = DC
ADB = ADC = 900
AD = AD (COMMON )
▲ADB = ▲ ADC (BY SAS RULE )
AB = AC (BY CPCT) A C D B Fig. 2.5
Slide 18: THEOREM 2. In a isosceles triangle altitude from the vertex bisects the base .
EXAMPLE: (in fig. 2.6)
GIVEN: An isosceles triangle
AB = AC
To prove : D bisects BC i.e. BD = DC
Proof:
IN ▲ ADB & ▲ADC
ADB = ADC
AD = AD
B = C ( AB = AC ; B = C)
▲ADB = ▲ ADC
BD = DC (BY CPCT) A C D B Fig. 2.6
Slide 19: THEOREM 3. If the bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is isosceles. EXAMPLE: (in fig. 2.7)
GIVEN: A ▲ABC in which AD bisects A meeting BC in D such that BD = DC, AD = DE
To prove : ▲ABC is isosceles triangle .
Proof: In ▲ ADB & ▲ EDC
BD = DC
AD = DE
ADB = EDC
▲ADB = ▲EDC
AB = EC
BAD = CED (BY CPCT)
BAD = CAD (GIVEN)
CAD = CED
AC = EC (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL)
AC = AB , HENCE ▲ABC IS AN ISOSCELES TRIANGLE. E D C B A Fig. 2.7
Slide 20: EXAMPLES QUES. IN FIG. AB = AC, D is the point in interior of ▲ABC such that
DBC = DCB. Prove that AD bisects BAC of ▲ ABC
SOLUTION: IN ▲BDC,
DBC = DCB (GIVEN)
DC = DB (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL.)
IN ▲ ABD & ▲ ACD
AB = AC (GIVEN)
BD = CD (GIVEN)
AD = AD (COMMON)
▲ABD = ▲ACD (BY SSS RULE)
BAD = CAD (BY CPCT)
Hence, AD is the bisector of ▲BAC A D B C
Slide 21: QUES. In fig. PA AB , QB AB & PA = QB.If PQ intersects AB at O, show that o is the mid point of AB as well as that of PQ
SOLUTION: In ▲OAP & ▲OBQ
PA = QB (GIVEN)
PAO = OBQ = 900 & AOP = BOQ
▲ OAP = ▲OBQ (BY AAS RULE)
OA = OB ( BY CPCT)
OP = OQ ( BY CPCT) O B A P Q
Slide 22: QUES. In fig. PS = PR, TPS = QPR.Prove that PT = PQ
SOLUTION: IN ▲PRS
PS = PR
PRS = PSR (ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL)
1800 - PRS = 1800 - PSR
PRQ = PST
Thus , in ▲ PST & ▲PRQ
TPS = QTR
PS = PR
PST = PRQ
▲PRT = ▲PSQ ( BY ASA RULE)
PT = PQ (BY CPCT) T S R Q P
Slide 23: QUES. In fig. BM and DN are both perpendicular to the line segment AC and BM = DN. Prove that AC bisects BD.
SOLUTION IN ▲BMR and ▲DNR
BMR = DNR (GIVEN)
BRM = DRN (VERTICALLY OPPOSITE ANGLES)
BM = DN (GIVEN)
▲ BMR = ▲DNR (BY AAS RULE)
BR = DR
R is the mid point of BD
Hence AC bisects BD B A C D M N R
Slide 24: QUES. In fig . It is given that BC = CE and 1 = 2. Prove that
▲ GCB = ▲ DCE
SOLUTION:
IN ▲ GCB & ▲ DCE
1 + GBC = 1800 (LINEAR PAIR )
2 + DEC = 1800 ( LINEAR PAIR)
1 + GBC = 2 + DEC -- (1) GBC = DEC (From (1) )
GBC = DEC (GIVEN)
BC = CE (GIVEN)
GCB = DCE (VERTICALLY OPPOSITE ANGLES )
▲GCB = ▲ DCE (BY ASA RULE)
*----- *---- *------ * -----* ----- * ----* ---- * A B C F G E D 1 2
Slide 25: MADE BY :- DEEPAK SHARMA CLASS :- IX - B SCHOOL = KAPIL GYANPEETH SUBMITTED TO : SURESH SIR