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PRESENTATION ON TRIANGLES NAME :- DEEPAK SHARMA CLASS:- IX C

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DESTINATION FOUND ……

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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

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CLASSIFICATION ON THE BASIS OF SIDES SCALENE TRIANGLE – If all the sides of triangle are unequal then it is a scalene triangle. 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 90 0 then it is an acute angled triangle. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 90 0 then it is a right angled triangle. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 90 0 then it is a obtuse angled triangle.

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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.

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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 60 0 60 0 A B C D F E Fig. 1.3

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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

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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

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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 = 90 0 , AC = DF , AB = DE TO PROVE : ▲ABC = ▲DEF B = E = 90 0 (GIVEN) AC = DF (GIVEN) AB = DE (GIVEN) ▲ABC = ▲DEF (BY RHS RULE) D F A B C 90 0 90 0 Fig. 1.7 E

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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

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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

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INEQUALITIES When two quantities are unequal then on comparing these quantities we obtain a relation between their measures called “ inequality “ relation.

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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

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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

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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

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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 = 90 0 THEN CDE < CED CD < CE ( Greater angles have larger side opposite to them. ) B A C E D Fig. 2.4

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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

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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

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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

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