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Areas of Similar Triangles

In two similar triangles, the ratio of the areas of the triangles is equal to the ratio of the squares of the corresponding sides.

Given

Δ
ABC ~ ΔDEF

To Prove    


Construction   
Draw AG
BC and DH EF.



Proof
                          
                    =            (i)

Now in the ΔABG and the ΔDEH
     B =E
  ∠AGB =DHE      (Each 90o)                   
ΔAGB ~ΔDHE


But, (
ΔABC ~ ΔDEF)
                     (ii)

From the Equation (i) and (ii), we get


Similarly, we can prove


Hence, 

 

Example

The area of two similar triangles are in the ratio of the squares of the corresponding altitudes.

Solution

Given
Two similar triangles ΔABC and ΔDEF where AL  BC and DM  EF. Also B E,  A D  
and
C F.


 

 

 

To Prove  


Proof   
Since 

ΔABC ~ ΔDEF
                                     (i)

Now in
ΔABL and ΔDEM,
     B =    E                                      (given)
   ALB = DME                                   (each 90o)
ΔABL ~ ΔDEM                                    (AA similarity)

Hence,                 
                        (ii)

From Equation (i) and (ii) we get,
  

 

Example
The areas of two similar triangles are in the ratio of the squares of the corresponding medians.


Given   
ΔABC ~ ΔDEF, and DL, AM are medians.
 

 

 

 


To Prove    


Proof   
Since 
ΔABC ~ ΔDEF

Therefore, 
(i) 

Also  

 

 (ii)

and B =E

ΔABM ~ ΔDEL

Hence,  (iii)     

 

Thus, from Equations (ii) and (iii), we get

     

(iv)
Hence, from Equations (i) and (iv), we get


Example
P, Q are points on sides AB, AC respectively in
ΔABC and PQ çç BC.  Prove that the median AD bisects PQ.

Solution



Given  
In the
ΔABC, AD is the median and PQ çç BC.

To Prove  
PR = QR

Proof  
In
ΔAPR and ΔABD
BAD =PAR    (Common angles)
ARP =ADB (corresponding angle of || sides)
APR =ABD
ΔAPR ~ ΔABD
 (i) Similarly,    (ii)

Comparing Equation (i) and (ii), we get,


But  BD = CD [D is the mid-point of BC]

PR = QR

Thus the median AD bisects PQ.


Example
If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar.


Solution
Given

 

 




 
 
Two triangles ABC and PQR in which AD and PS are median such that  

To Prove   

ΔABC ~ ΔPQR

Construction   
Draw DE
çç AB and ST çç PQ

Proof   
In the
ΔABC, D is the mid-point of BC and DE çç AB,

DE= AB and E will be mid-point of AC.

Similarly, ST=
PQ and T is mid-point of PR

We are given,


                 
or     


        ΔADE ~ ΔPST

        DAE = SPT          (i)

Similarly we can prove,
          

           BAD = QPS            (ii)

Adding Equations (i) and (ii) we get,


            BAD + DAE = SPT + QPS

          BAD + DAC = SPR + QPS

 

    BAC = QPR

Also we are given

ΔABC ~ Δ PQR [SAS similarity of triangles]


Example
P is the mid-point of side BC of
ΔABC.  If Q is the mid-point of AP and BQ when produced meets AC at L, prove that AL =AC. 
Given  

D
ABC, in which P is the mid-point of BC and Q is the mid-point of AP
 


To Prove   
AL =AC

Construction   
Draw PO çç BL meeting AC in O.

Proof   
In the
ΔBCL, PO çç BL and P is the mid-point of BC.

... O is the mid-point of CL.  Now in the
ΔAPO, QL çç PO and Q is the mid-point of AP, 

... L is the mid-point of AO

AL = LO = OC

and   AL + LO + OC = AC


  AL + AL + AL = AC

                or  AL= AC

 

Example
If the areas of two similar triangles are equal, then the two triangles will be congruent.

 



Solution
Given   

ΔABC ~ ΔDEF and ar (ΔABC) = ar (ΔDEF)

To Prove      

    DABC≅ΔDEF


Proof 

 

    

 

 

        1=
   

AB = DE, BC = EF and AC = DF

ΔABC ≅ΔDEF [SSS Congruence]  
  
                                                     

Example
In the given figure, find x if AB çç DC.
 

 

Solution
In the quadrilateral ABCD, AB || DC.


ABD = BDC and CAB = ACD

ΔABO ~ ΔCDO

                  

                  

or      3(3x - 19) = (x - 5)(x - 3)

or   x2 - 17x + 72 = 0

or   (x - 9)(x - 8)=0


         x = 9 or 8
   

Example
In the fig
. AB || CD || EF.Prove that 

 



Solution
In
ΔBCD, EF || CD

 

     .....(1)   

In ΔABD, EF çç AB



1 -

 

 

1 -                            (from 1)

1 =

 

 

1 = EF

 

 

or  


Example
In the fig. DEFG is a square and
A = 90°

Prove that DE
2 = BD ´ EC.
 


Solution
In
ΔAGF and ΔDBG

GAF =BDG  (each 90°)

 ∠AGF = GBD    (Corresponding angles of  çç sides)

ΔAFG ~ΔDBG   (By AA corollary)

Similarly, 
   ΔAGF ~ ΔEFC
...                ΔBDG ~ ΔFEC

 

Hence, 

[DE = EF = FG = GD]

Hence,
DE2 = BD x EC    

Example
In a ||
gm ABCD, the diagonal BD intersects the segment AE at F where E is the mid-point on BC, prove that DF = 2FB

In the
|| gm ABCD (fig). DC çç AB and DA çç CB.

 

 


Solution
In Δ ADF and ΔBEF
 

       ADF = EBF  (alternate angles)

       DFA = EFB (Vertically Opposite angles)
                  
...  
ΔA DF ~ ΔEBF

...  


...  

or  = 2


Hence, DF = 2 FB.

 

 


 





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