Triangle class 10

Triangle class 10

Congruent figures : Two geometric figures which have the same shape and size are known as congruent figures.

Similar figures : geometric figures which have same shape but different sizes are known as similar figures.

Two congruent figures are always similar but two similar figures need not be congruent.

SIMILAR POLYGONS : Two polygons having the same number of sides are said to be similar, if 

(i) their corresponding angles are equal, and 

(ii) the lengths of their corresponding sides are proportional.

if two polygons ABCDE and PQRST are similar, we write, ABCDE ~ PQRST , where the symbol ‘~’ stands for similar to’

The constant ratio between the corresponding sides of two similar figures is known as the scale factor, or the representative fraction. Since triangles are also polygons, so the same set of condition apply for the similarity of triangles.

EQUIANGULAR TRIANGLES : Two triangles are said to be equiangular if their corresponding angles are equal.

SIMILAR TRISNGLES : Two triangle are said to be similar to each other if,

(i) their corresponding angles are equal, and 

(ii) their corresponding sides are proportional. 

Basic proportionality theorem (b.p.t) or Thales theorem

*)Thales' Theorem

Here Thales’ theorem is describe by two method and named as method – 1 and method – 2. and for video of these theory click on the respective method 

please read the below pdf in which Thales’ theorem has been describe 

Method -1st

Method -2nd

*)converse of Thales' Theorem video lecture :-

Converse of thales' theorem

we have already learned the two mothed of  Thales’ theorem in detailed way. each and every point have been discussed in that two method if you have not seen or gone through that lecture then please visit here for video lecture method -1 and method -2 and click here for notes 

following is the brief of  Converse of thales’ theorem  

mathematics class 10

CRITERIA FOR SIMILARITY OF TWO TRIANGLES

(AAA-SIMILARITY) : If in two triangle, the corresponding angle are equal, then their corresponding sides are proportional and hence the triangle are similar.

Given :- \Delta ABC and \Delta DPQ, WE have

To prove : \Delta ABC ~ \Delta  

Construction : Cut DP = AB and DQ = AC . join PQ

AB = DP [by construction]

AC = DQ [by construction]

\angle A = \angle D [given]

\therefore   \Delta ABC \cong \Delta DPQ [by SAS-congruence]

\angle B = \angle p

\angle E = \angle p [ \because \angle B = \angle E (given)] 

PQ || EF [ \because corresponding \angle s are equal ]

\dfrac{DP}{DE} = \dfrac{DQ}{DF}

\dfrac{AB}{DE} = \dfrac{CA}{FD} [ \because DP = AB and DQ = AC ]

Similarly , \dfrac{AB}{DE} = \dfrac{BC}{EF} . 

\therefore \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD} .

Thus, \angle A = \angle D , \angle B = \angle E , \angle C = \angle F and \dfrac{ AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD} .

Hence, \Delta  ABC \sim \Delta DEF.

( AA-SIMILARITY ) : If two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar.

Proof : In \Delta ABC = \Delta DEF , let \angle A = \Delta D  and  \angle B = \angle E .

Then, 3rd \angle C   = 3rd \angle F .

AA-similarity is the same as AAA-similarity.

SSS-SIMILARITY : If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two tringles are similar.

Given :-  \Delta  ABC   and   \Delta DEF   in   which   \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} .

Prove :- \Delta ABC \sim \Delta DEF .

Construction :- Let us take \Delta ABC   and   \Delta DEF such that

\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}     (<1).

Cut DP = AB and DQ = AC. Join PQ.

Proof :- \dfrac{AB}{DE} = \dfrac{AC}{DF} \Rightarrow  \dfrac{DP}{DE} = \dfrac{DQ}{DF} [ \because AB = DP    and    AC =DQ]

So, by the converse of thales’ theorem , PQ || EF.

\therefore  \angle P = \angle E     [corresponding    \angle s ]

\angle Q = \angle F     [corresponding    \angle s ]

\therefore  \Delta DPQ \sim \Delta DEF        [by AA-similarity]   

\Rightarrow  \dfrac{DP}{DE} = \dfrac{PQ}{EF}

\Rightarrow  \dfrac{AB}{DE} = \dfrac{PQ}{EF}.  …..(i) [ \because DP = AB ]

But \dfrac{AB}{DE} = \dfrac{BC}{EF}.  …..(ii) [given]

\because  \dfrac{PQ}{EF} = \dfrac{BC}{EF}  [from(i) and (ii) ]

\Rightarrow  BC = PQ .

Thus, AB = DP, AC = DQ and BC = PQ.

\therefore  \Delta ABC \cong \Delta DPQ     [by SSS – similarity ]   

\therefore  \angle A = \angle D , \angle B = \angle P = \angle and \angle C = \angle Q =\angle F

\Rightarrow  \angle A = \angle D , \angle B = \angle E and  \angle C = \angle F .

Thus, the given triangle are equiangular and hence similar.

SAS-SIMILARITY  :- If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangle are similar.
Given :- \Delta ABC   and   \Delta DEF   in   which   \angle A = \angle D   and   \dfrac{AB}{DE} = \dfrac{AC}{DF} 
To prove :- \Delta ABC \sim \Delta DEF.
CONSTRUCTION:- Let us take \Delta ABC   and   \Delta DEF   such    that   
\dfrac{AB}{DE} = \dfrac{AC}{DF}  (<1)   and   \angle A   =   \angle D.
Cut DP = AB and DQ =AC. Join PQ.
Proof :- In \Delta ABC   and   \Delta DPQ  ,   we   have
AB = DP [ by construction ]
\angle A = \angle D [given]
AC = DQ [by construction]
\therefore  \Delta ABC \cong \Delta DPQ [by   SAS  –  congruence] 
\therefore \angle A = \angle D , \angle B = \angle P and \angle C = \angle Q.
Now, \dfrac{AB}{DE} = \dfrac{AC}{DF} (given)
\Rightarrow  \dfrac{DP}{DE} = \dfrac{DQ}{DF}      [\because AB = DP and AC = DQ ]
\Rightarrow  PQ || EF    [ By  the  converse   of   thales'   theoram ]
\Rightarrow  \angle P = \angle E   and   \angle Q = \angle F   [ corresponding   \angle s ]
\therefore  \angle A =  \angle D , \angle B = \angle P = \angle E   and   \angle C = \angle Q = \angle F.
Thus,   \angle A = \angle D , \angle B = \angle E , \ angle C = \angle F.
So, the given triangle are equiangular and hence similar.
Theorem :- If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse the the triangle on the triangle on the both sides of the perpendiculer are similar to the whole triangle and also to each other
Given :- A \Delta ABC   in   which   \angle BAC = 90\degree   and   AD \perp BC
To prove (i) \Delta DBA \sim \Delta ABC
(ii)   \Delta DAC \sim \Delta ABC
(iii)   \Delta DBA \sim \Delta DAC
Proof :- (i) In \Delta DBA   and   \Delta ABC, we have
\angle BDA = \angle BAC = 90 \degree
\angle DBA = \angle   ABC   (common)   
\therefore  \Delta DBA \sim  \Delta ABC  [by AA- similarity ]
(ii) In \Delta DAC   and   \Delta ABC ,  we    have
\angle CDA = \angle CAB = 90 \degree
\angle DCA = \angle ACB   (common)   
\therefore  \Delta DAC \sim \Delta ABC   [ By    AA-   similarity ].
(iii)  In   \Delta DBA   and  \Delta DAC, we have
\angle ADB  =  \angle CDA = 90 \degree    [/latex]
\therefore  \angle B + \angle BAD = 90\degree   …(1)
\angle C  +  \angle CAD = 90 \degree   …(2)
  \angle BAD  +  \angle CAD  = 90 \degree …(3)
form (1) ,(2) and (3)
  \Rightarrow  \angle B  = \angle  CAD   and   \ C   and   \angle BAD .
Thus, in   \Delta DBA   and   \ DAC   we   have   :   
\angle ADB = \angle CDA  [each  equal  to   90 \degree ]
  \angle B = \angle CAD 
\angle BAD = \angle C
\therefore  \Delta DBA \sim \Delta DAC  [ by AAA – similarity ]

Theorem : – If two triangle are equiangular , prove that the ratio of their corresponding sides is the same as the ratio of the corresponding altitudes . 

\Delta ABC and \Delta DEF  in  which \angle A = \angle D , \angle B = \angle E and \angle C = \angle F and  AF \perp BC  and  DM \perp EF.

TO PROVE :   \dfrac{BC}{EF}  =  \dfrac{AL}{DM}

 

PROOF :  Since   \Delta ABC  and  \Delta  DEF  are equiangular , \Delta ABC  \sim  \Delta DEF.  

 

  \therefore \dfrac{AB}{DE}  =  \dfrac{BC}{EF}

 

In \Delta ALB  and  \Delta DME , we have  

 

  \angle ALB  =  \angle DME  =  90\degree   and  \angle B  = \angle E (given) . 

 

\therefore  \Delta ALB  \sim  \Delta  DME  [ by AA – SIMILARITY ]

 

\therefore  \dfrac{AB}{DE} = \dfrac{AL}{DM}.

 

form (i) and (ii) , we get  \dfrac{BC}{EF}  =  \dfrac{AL}{DM}  

Theorem : If two triangle are equiangular  , prove that the ratio of their corresponding sides is the same as the ratio of the  corresponding medians. 

GIVEN :  \Delta ABC  and  \Delta  DEF  in which \angle A  =  \angle D , \angle B = \angle E and \angle C = \angle F  and  AL   and  DM  are the medians

TO PROVE  :  \dfrac{BC}{EF}  =  \dfrac{AL}{DM}  

PROOF : Since \Delta ABC  and  \Delta DEF  are   equiangular  ,  we  have  \Delta ABC   \sim  \Delta DEF  

  \therefore  \dfrac{AB}{DE}  =  \dfrac{BC}{EF}. …(i)

But,   \dfrac{AB}{DE}  =  \dfrac{BC}{EF}  = \dfrac{2BL}{2EM}  = \dfrac{BL}{EM}.

Now , in   \Delta ABL  and  \Delta  DEM   , we  have \dfrac{AB}{DE}  =  \dfrac{BL}{EM}  and   \angle B  =  \angle E (given).

  \therefore  \Delta ABL  \sim  \Delta DEM   [by   SAS –  similarity ]

\Rightarrow  \dfrac{AB}{DE}  =  \dfrac{AL}{DM} . …..(ii) 

From (i) and (ii) , we get  \dfrac{BC}{EF}  =  \dfrac{AL}{DM}.

Theorem :  If two triangle are equiangular , show that the ratio of the corresponding sides is the same as the ratio of the corresponding angle – bisector segments.

Given : \Delta ABC   and   \Delta DEF   in  which   \angle A  =  \angle D , \angle B  =  \angle E   and   \angle C  = \angle F ,  and  AX   and  DY  are  the   bisectors  of   \angle A  and   \angle D   respectively  .   

TO Prove :   \dfrac{BC}{EF}  =  \dfrac{AX}{DY} 

Proof : Since    \Delta ABC   and   \Delta DEF   are   equiangular,   we   have   \Delta ABC  \sim  \Delta DEF .  

\therefore  \dfrac{AB}{DE}   =  \dfrac{BC}{EF}   …..  (i) 

Now,  \angle A  =  \angle D  \Rightarrow  \dfrac{1}{2}   \angle A  =  \dfrac{1}{2}  \angle D  \Rightarrow   \angle BAX  =  \angle EDY .

Thus, in   \Delta ABX    and  \Delta DEY ,  we   have   

  \angle BAX  =  \angle EDY    (proved)    \angle B  =  \angle E (given) \therefore  \Delta ABX  \sim  \Delta DEY    [ by   AA  –  similarity ] \Rightarrow  \dfrac{AB}{DE}  =  \dfrac{AX}{DY} . ….(ii)

From (i) and (ii) , we get  \dfrac{BC}{EF}  =  \dfrac{AX}{DY}.  

THEOREM  : The ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides. 

GIVEN :  \Delta ABC  \sim  \Delta DEF .  

TO PROVE : \dfrac{ar \Delta ABC}{ar \Delta DEF}  = \dfrac{AB^2}{DE^2} = \dfrac{AC^2}{DF^2} = \dfrac{BC^2}{EF^2} 

CONSTRUCTION : Draw AL \perp BC  and  DM \perp  EF . .

PROOF : Since \Delta ABC  \sim  \Delta DEF , it follows that they are equiangular and their side are proportional . 

\therefore  \angle A  =  \angle D ,  \angle B =  \angle E ,  \angle C  =  \angle F  and  \dfrac{AB}{DE}  =  \dfrac{BC}{EF}  = \dfrac{AC}{DF}. …..(i)

Now, ar (\Delta ABC )  =  (\dfrac{1}{2} \times BC \times  AL )  

and , ar (\Delta DEF )  =  (\dfrac{1}{2}  \times  EF  \times DM ). 

\therefore  ( \dfrac{ar \Delta ABC}{ar \Delta DEF})  =  \dfrac{ \dfrac{1}{2}  \times  BC  \times  AL}{\dfrac{1}{2}  \times EF  \times  DM }  =  \dfrac{BC}{EF}  \times  \dfrac{AL}{DM}.  ….(ii)

Also, \dfrac{AL}{DM}  =  \dfrac{BC}{EF}  …(iii)  [\because  in similar triangles, the ratio of the corres . sides is the same as the ratio of corresponding altitudes ]

Using (iii) in (ii) , we get 

  \dfrac{ar \Delta ABC}{ar \Delta DEF}  =  (\dfrac{BC}{EF}  \times  \dfrac{BC}{EF})  =  \dfrac{BC^2}{EF^2}

 

Similarly, \dfrac{ar \Delta ABC}{ar \Delta DEF}  =  \dfrac{AB^2}{DE^2}  and  \dfrac{ar \Delta ABC}{ar \Delta DEF}  =  \dfrac{AC^2}{DF^2}  

Hence, \dfrac{ar \Delta ABC}{ar \Delta DEF}  =  \dfrac{AB^2}{DE^2}  =  \dfrac{AC^2}{DF^2}  =  \dfrac{BC^2}{EF^2}

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